PROJECT : MAA BHARTI NURSING COLLEGE, GAURAVPATH, KUSHALGARH
PLAN : nurse
JOB NO.: 1 REF. NO.:
DATE: 11/06/25 TIME: 09:43:38
Building Version: 1.640 STRUDS Version: 4.0.0


Terms Used In Calculation :
In biaxial column design,
Pu = Puc + Pus(Total)
               where,
               Pu=external axial compressive load,
               Puc=axial compressive resistance offered by concrete,
               Pus(Total)=total axial compressive resistance offered by steel
               at different levels in the section.
Mu = Muc + Mus(Total)
               where,
               Mu=external moment about centroidal axis,
               Muc=moment of resistance offered by concrete in compression,
               Mus(Total)=total moment of resistance offered by steel
              at different levels in the section.
i= serial no. of row of reinforcement,
Asi= cross-sectional area of reinforcement in the i th row,
fsi= stress in the reinforcement in the i th row (compression + ve, tension - ve),
fci= compressive stress in concrete at the level of i th row of reinforcement,
ei= strain at the i th row of reinforcement from the stress-strain curve of steel and concrete,
xi= distance of the bars in the i th row from the centroid of the section
Values of stress in steel :
Design strength in bending compression (fyd) = 0.87 fy
              (a) Fe 250,
               For ei >= fyd / Es, fsi = fyd
               For ei < fyd / Es, fsi = ei x fyd
              (b) Fe 415,
               For ei >= 0.8 x fyd / Es, fsi = value obtained from stress-strain curve
               For ei < 0.8 x fyd / Es, fsi = ei x fyd
Stress Stress level
(N/mm˛)
Strain
0.800 fyd 288.70.00144
0.850 fyd 306.70.00163
0.900 fyd 324.80.00192
0.950 fyd 342.80.00241
0.975 fyd 351.80.00276
1.000 fyd 360.90.00380
Values of stress in concrete :
              For ei >= 0.002, fci = 0.446 fck
              For ei < 0.002, fci = (446.ei x (1 - 250.ei)) x fck
Values of strain :
               Values of strain at different levels are obtained by taking maximum strain
               as 0.0035 as the reference value at the highly compressed edge.


Design of CG1 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 199.50 kN
MomentX,(Mx) = 37.38 kN-m
MomentY,(My) = 25.27 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C90.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 4.130 kN.m
My_MinEccen = 3.990 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 37.384 kN.m
My = max(My,My_MinEccen) + MuaddY = 25.273 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 37.38/199.50 = 187 mm
Actual eccenY = My / P = 25.27/199.50 = 127 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(187,21) = 187 mm
eccenY = max(Actual eccenY,eccenYMin) = max(127,20) = 127 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(187 mm) > 0.05 x 300(15 mm)
and eccenY(127 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00247 344.22 8.92 335.30 75.84 179 13.58
2 226 0.00045 90.29 3.57 86.72 19.62 90 1.76
3 226 -0.00156 -300.27 0.00 -300.27 -67.92 0 0.00
4 226 -0.00358 -358.94 0.00 -358.94 -81.19 -90 7.27
5 226 -0.00559 -360.90 0.00 -360.90 -81.63 -179 14.61
Total-135.29 37.21
xux = 156 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 156/450 = 0.124
Puc = 0.124 x 20.00 x 300 x 450
= 336.02 kN
Pux1 = Puc + Pus(Total)
= 336.02 + (-135.29)
= 200.74 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 156/450
= 0.144
Muc = 336.02 x (0.5 x 450 - 0.144 x 450) = 53.86 kN-m
Mux1 = Muc + Mus(Total)
= 53.86 + (37.21)
= 91.07 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00141 282.09 8.14 273.95 154.91 104 16.11
2 565 -0.00804 -360.90 0.00 -360.90 -204.08 -104 21.22
Total-49.17 37.34
xuy = 77 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 77/300 = 0.092
Puc = 0.092 x 20.00 x 450 x 300
= 249.64 kN
Puy1 = Puc + Pus(Total)
= 249.64 + (-49.17)
= 200.47 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 77/300
= 0.107
Muc = 249.64 x (0.5 x 300 - 0.107 x 300) = 29.44 kN-m
Muy1 = Muc + Mus(Total)
= 29.44 + (37.34)
= 66.78 kN-m
Pu/Puz = 0.128
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.128, an = 1.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((37.38/91.07)1.000) + ((25.27/66.78)1.000)
= 0.789 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG1 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 419.82 kN
MomentX,(Mx) = 32.28 kN-m
MomentY,(My) = 23.17 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C90.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 8.690 kN.m
My_MinEccen = 8.396 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 32.275 kN.m
My = max(My,My_MinEccen) + MuaddY = 23.170 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 32.28/419.82 = 77 mm
Actual eccenY = My / P = 23.17/419.82 = 55 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(77,21) = 77 mm
eccenY = max(Actual eccenY,eccenYMin) = max(55,20) = 55 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(77 mm) > 0.05 x 300(15 mm)
and eccenY(55 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00273 351.12 8.92 342.20 77.40 179 13.86
2 226 0.00124 248.46 7.64 240.82 54.47 90 4.88
3 226 -0.00025 -49.79 0.00 -49.79 -11.26 0 0.00
4 226 -0.00174 -313.58 0.00 -313.58 -70.93 -90 6.35
5 226 -0.00323 -355.93 0.00 -355.93 -80.51 -179 14.41
Total-30.82 39.49
xux = 210 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 210/450 = 0.168
Puc = 0.168 x 20.00 x 300 x 450
= 453.73 kN
Pux1 = Puc + Pus(Total)
= 453.73 + (-30.82)
= 422.90 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 210/450
= 0.194
Muc = 453.73 x (0.5 x 450 - 0.194 x 450) = 62.44 kN-m
Mux1 = Muc + Mus(Total)
= 62.44 + (39.49)
= 101.93 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00231 338.96 8.92 330.04 186.63 104 19.41
2 565 -0.00310 -354.75 0.00 -354.75 -200.60 -104 20.86
Total-13.97 40.27
xuy = 135 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 135/300 = 0.162
Puc = 0.162 x 20.00 x 450 x 300
= 436.64 kN
Puy1 = Puc + Pus(Total)
= 436.64 + (-13.97)
= 422.67 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 135/300
= 0.187
Muc = 436.64 x (0.5 x 300 - 0.187 x 300) = 41.02 kN-m
Muy1 = Muc + Mus(Total)
= 41.02 + (40.27)
= 81.29 kN-m
Pu/Puz = 0.270
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.270, an = 1.116
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((32.28/101.93)1.116) + ((23.17/81.29)1.116)
= 0.523 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG1 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 628.56 kN
MomentX,(Mx) = 29.11 kN-m
MomentY,(My) = 20.90 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C90.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 13.011 kN.m
My_MinEccen = 12.571 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 29.108 kN.m
My = max(My,My_MinEccen) + MuaddY = 20.904 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 29.11/628.56 = 46 mm
Actual eccenY = My / P = 20.90/628.56 = 33 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(46,21) = 46 mm
eccenY = max(Actual eccenY,eccenYMin) = max(33,20) = 33 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(46 mm) > 0.05 x 300(15 mm)
and eccenY(33 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00289 352.91 8.92 343.99 77.81 179 13.93
2 226 0.00170 310.78 8.71 302.07 68.33 90 6.12
3 226 0.00050 100.67 3.92 96.74 21.88 0 0.00
4 226 -0.00069 -137.73 0.00 -137.73 -31.15 -90 2.79
5 226 -0.00188 -322.34 0.00 -322.34 -72.91 -179 13.05
Total63.95 35.88
xux = 263 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 263/450 = 0.210
Puc = 0.210 x 20.00 x 300 x 450
= 567.63 kN
Pux1 = Puc + Pus(Total)
= 567.63 + (63.95)
= 631.58 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 263/450
= 0.243
Muc = 567.63 x (0.5 x 450 - 0.243 x 450) = 65.66 kN-m
Mux1 = Muc + Mus(Total)
= 65.66 + (35.88)
= 101.55 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00262 348.18 8.92 339.26 191.85 104 19.95
2 565 -0.00136 -272.58 0.00 -272.58 -154.14 -104 16.03
Total37.71 35.98
xuy = 183 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 183/300 = 0.219
Puc = 0.219 x 20.00 x 450 x 300
= 592.31 kN
Puy1 = Puc + Pus(Total)
= 592.31 + (37.71)
= 630.02 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 183/300
= 0.254
Muc = 592.31 x (0.5 x 300 - 0.254 x 300) = 43.80 kN-m
Muy1 = Muc + Mus(Total)
= 43.80 + (35.98)
= 79.78 kN-m
Pu/Puz = 0.404
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.404, an = 1.340
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((29.11/101.55)1.340) + ((20.90/79.78)1.340)
= 0.354 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG1 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 819.99 kN
MomentX,(Mx) = 24.11 kN-m
MomentY,(My) = 17.12 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C90.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 16.974 kN.m
My_MinEccen = 16.400 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 24.106 kN.m
My = max(My,My_MinEccen) + MuaddY = 17.119 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 24.11/819.99 = 29 mm
Actual eccenY = My / P = 17.12/819.99 = 21 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(29,21) = 29 mm
eccenY = max(Actual eccenY,eccenYMin) = max(21,20) = 21 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(29 mm) > 0.05 x 300(15 mm)
and eccenY(21 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00299 353.77 8.92 344.85 78.00 179 13.96
2 226 0.00198 327.16 8.92 318.24 71.98 90 6.44
3 226 0.00098 196.63 6.61 190.02 42.98 0 0.00
4 226 -0.00002 -3.60 0.00 -3.60 -0.81 -90 0.07
5 226 -0.00102 -203.83 0.00 -203.83 -46.11 -179 8.25
Total146.05 28.73
xux = 313 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 313/450 = 0.250
Puc = 0.250 x 20.00 x 300 x 450
= 675.84 kN
Pux1 = Puc + Pus(Total)
= 675.84 + (146.05)
= 821.89 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 313/450
= 0.289
Muc = 675.84 x (0.5 x 450 - 0.289 x 450) = 64.10 kN-m
Mux1 = Muc + Mus(Total)
= 64.10 + (28.73)
= 92.83 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00275 351.63 8.92 342.71 193.80 104 20.15
2 565 -0.00062 -124.58 0.00 -124.58 -70.45 -104 7.33
Total123.35 27.48
xuy = 216 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 216/300 = 0.259
Puc = 0.259 x 20.00 x 450 x 300
= 698.63 kN
Puy1 = Puc + Pus(Total)
= 698.63 + (123.35)
= 821.97 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 216/300
= 0.299
Muc = 698.63 x (0.5 x 300 - 0.299 x 300) = 42.13 kN-m
Muy1 = Muc + Mus(Total)
= 42.13 + (27.48)
= 69.61 kN-m
Pu/Puz = 0.527
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.527, an = 1.545
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((24.11/92.83)1.545) + ((17.12/69.61)1.545)
= 0.239 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG1 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 983.05 kN
MomentX,(Mx) = 14.03 kN-m
MomentY,(My) = 10.07 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C90.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 21.381 kN.m
My_MinEccen = 19.661 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 21.381 kN.m
My = max(My,My_MinEccen) + MuaddY = 19.661 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 14.03/983.05 = 14 mm
Actual eccenY = My / P = 10.07/983.05 = 10 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(14,22) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(10,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00305 354.37 8.92 345.45 78.14 179 13.99
2 226 0.00218 334.50 8.92 325.58 73.64 90 6.59
3 226 0.00131 262.93 7.87 255.05 57.69 0 0.00
4 226 0.00045 89.07 3.53 85.54 19.35 -90 -1.73
5 226 -0.00042 -84.79 0.00 -84.79 -19.18 -179 3.43
Total209.64 22.28
xux = 360 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 360/450 = 0.288
Puc = 0.288 x 20.00 x 300 x 450
= 778.36 kN
Pux1 = Puc + Pus(Total)
= 778.36 + (209.64)
= 988.00 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 360/450
= 0.333
Muc = 778.36 x (0.5 x 450 - 0.333 x 450) = 58.45 kN-m
Mux1 = Muc + Mus(Total)
= 58.45 + (22.28)
= 80.73 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00285 352.60 8.92 343.68 194.35 104 20.21
2 565 -0.00008 -15.67 0.00 -15.67 -8.86 -104 0.92
Total185.49 21.13
xuy = 248 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 248/300 = 0.298
Puc = 0.298 x 20.00 x 450 x 300
= 804.94 kN
Puy1 = Puc + Pus(Total)
= 804.94 + (185.49)
= 990.42 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 248/300
= 0.345
Muc = 804.94 x (0.5 x 300 - 0.345 x 300) = 37.55 kN-m
Muy1 = Muc + Mus(Total)
= 37.55 + (21.13)
= 58.68 kN-m
Pu/Puz = 0.631
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.631, an = 1.719
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((21.38/80.73)1.719) + ((19.66/58.68)1.719)
= 0.254 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG1 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG2 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 291.16 kN
MomentX,(Mx) = 103.69 kN-m
MomentY,(My) = 3.84 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C100.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 6.027 kN.m
My_MinEccen = 5.823 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 103.693 kN.m
My = max(My,My_MinEccen) + MuaddY = 5.823 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 103.69/291.16 = 356 mm
Actual eccenY = My / P = 3.84/291.16 = 13 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(356,21) = 356 mm
eccenY = max(Actual eccenY,eccenYMin) = max(13,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(356 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 2 nos. + #16 - 6 nos. (1433 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00261 347.97 8.92 339.05 136.34 177 24.13
2 226 0.00044 87.54 3.48 84.06 19.01 60 1.13
3 402 -0.00174 -313.29 0.00 -313.29 -125.98 -58 7.26
4 402 -0.00395 -360.90 0.00 -360.90 -145.13 -177 25.69
Total-115.75 58.22
xux = 189 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 189/450 = 0.151
Puc = 0.151 x 20.00 x 300 x 450
= 408.16 kN
Pux1 = Puc + Pus(Total)
= 408.16 + (-115.75)
= 292.41 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 189/450
= 0.175
Muc = 408.16 x (0.5 x 450 - 0.175 x 450) = 59.75 kN-m
Mux1 = Muc + Mus(Total)
= 59.75 + (58.22)
= 117.97 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 716 0.00184 319.97 8.86 311.11 222.84 102 22.73
2 716 -0.00520 -360.90 0.00 -360.90 -258.51 -102 26.37
Total-35.67 49.10
xuy = 101 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 101/300 = 0.122
Puc = 0.122 x 20.00 x 450 x 300
= 328.43 kN
Puy1 = Puc + Pus(Total)
= 328.43 + (-35.67)
= 292.76 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 101/300
= 0.141
Muc = 328.43 x (0.5 x 300 - 0.141 x 300) = 35.41 kN-m
Muy1 = Muc + Mus(Total)
= 35.41 + (49.10)
= 84.51 kN-m
Pu/Puz = 0.177
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.177, an = 1.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((103.69/117.97)1.000) + ((5.82/84.51)1.000)
= 0.948 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG2 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 2 nos. + #16 - 6 nos. (1433 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 595.50 kN
MomentX,(Mx) = 93.64 kN-m
MomentY,(My) = 2.75 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C100.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 12.327 kN.m
My_MinEccen = 11.910 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 93.636 kN.m
My = max(My,My_MinEccen) + MuaddY = 11.910 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 93.64/595.50 = 157 mm
Actual eccenY = My / P = 2.75/595.50 = 5 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(157,21) = 157 mm
eccenY = max(Actual eccenY,eccenYMin) = max(5,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(157 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00286 352.71 8.92 343.79 77.76 179 13.92
2 226 0.00163 306.36 8.61 297.75 67.35 90 6.03
3 226 0.00039 77.78 3.13 74.65 16.88 0 0.00
4 226 -0.00085 -169.73 0.00 -169.73 -38.39 -90 3.44
5 226 -0.00209 -330.90 0.00 -330.90 -74.85 -179 13.40
Total48.76 36.78
xux = 253 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 253/450 = 0.203
Puc = 0.203 x 20.00 x 300 x 450
= 546.75 kN
Pux1 = Puc + Pus(Total)
= 546.75 + (48.76)
= 595.51 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 253/450
= 0.234
Muc = 546.75 x (0.5 x 450 - 0.234 x 450) = 65.45 kN-m
Mux1 = Muc + Mus(Total)
= 65.45 + (36.78)
= 102.23 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00259 347.43 8.92 338.51 191.42 104 19.91
2 565 -0.00152 -296.65 0.00 -296.65 -167.75 -104 17.45
Total23.67 37.35
xuy = 177 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 177/300 = 0.212
Puc = 0.212 x 20.00 x 450 x 300
= 573.33 kN
Puy1 = Puc + Pus(Total)
= 573.33 + (23.67)
= 597.00 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 177/300
= 0.245
Muc = 573.33 x (0.5 x 300 - 0.245 x 300) = 43.80 kN-m
Muy1 = Muc + Mus(Total)
= 43.80 + (37.35)
= 81.15 kN-m
Pu/Puz = 0.383
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.383, an = 1.304
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((93.64/102.23)1.304) + ((11.91/81.15)1.304)
= 0.974 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG2 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 916.58 kN
MomentX,(Mx) = 92.91 kN-m
MomentY,(My) = 1.85 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C100.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 18.973 kN.m
My_MinEccen = 18.332 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 92.914 kN.m
My = max(My,My_MinEccen) + MuaddY = 18.332 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 92.91/916.58 = 101 mm
Actual eccenY = My / P = 1.85/916.58 = 2 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(101,21) = 101 mm
eccenY = max(Actual eccenY,eccenYMin) = max(2,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(101 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 6 nos. + #16 - 4 nos. (1483 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00299 353.83 8.92 344.91 138.70 177 24.55
2 226 0.00204 329.35 8.92 320.43 72.48 88 6.34
3 226 0.00112 223.40 7.18 216.22 48.91 0 0.00
4 226 0.00019 38.06 1.62 36.44 8.24 -88 -0.72
5 402 -0.00076 -151.52 0.00 -151.52 -60.93 -177 10.78
Total207.40 40.95
xux = 330 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 330/450 = 0.264
Puc = 0.264 x 20.00 x 300 x 450
= 713.81 kN
Pux1 = Puc + Pus(Total)
= 713.81 + (207.40)
= 921.21 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 330/450
= 0.306
Muc = 713.81 x (0.5 x 450 - 0.306 x 450) = 62.48 kN-m
Mux1 = Muc + Mus(Total)
= 62.48 + (40.95)
= 103.43 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 741 0.00275 351.63 8.92 342.71 254.09 102 25.92
2 741 -0.00042 -84.00 0.00 -84.00 -62.28 -102 6.35
Total191.81 32.27
xuy = 225 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 225/300 = 0.270
Puc = 0.270 x 20.00 x 450 x 300
= 729.00 kN
Puy1 = Puc + Pus(Total)
= 729.00 + (191.81)
= 920.81 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 225/300
= 0.312
Muc = 729.00 x (0.5 x 300 - 0.312 x 300) = 41.12 kN-m
Muy1 = Muc + Mus(Total)
= 41.12 + (32.27)
= 73.39 kN-m
Pu/Puz = 0.551
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.551, an = 1.585
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((92.91/103.43)1.585) + ((18.33/73.39)1.585)
= 0.955 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG2 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 6 nos. + #16 - 4 nos. (1483 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1258.60 kN
MomentX,(Mx) = 91.36 kN-m
MomentY,(My) = 0.97 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C100.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 26.053 kN.m
My_MinEccen = 25.172 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 91.364 kN.m
My = max(My,My_MinEccen) + MuaddY = 25.172 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 91.36/1258.60 = 73 mm
Actual eccenY = My / P = 0.97/1258.60 = 1 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(73,21) = 73 mm
eccenY = max(Actual eccenY,eccenYMin) = max(1,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(73 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 12 nos. (2413 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00304 354.25 8.92 345.33 138.87 177 24.58
2 402 0.00236 341.07 8.92 332.14 133.56 106 14.18
3 402 0.00169 310.13 8.70 301.44 121.21 35 4.29
4 402 0.00101 201.46 6.72 194.73 78.31 -35 -2.77
5 402 0.00033 65.91 2.70 63.21 25.42 -106 -2.70
6 402 -0.00035 -69.64 0.00 -69.64 -28.00 -177 4.96
Total469.37 42.54
xux = 366 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 366/450 = 0.293
Puc = 0.293 x 20.00 x 300 x 450
= 789.75 kN
Pux1 = Puc + Pus(Total)
= 789.75 + (469.37)
= 1259.12 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 366/450
= 0.338
Muc = 789.75 x (0.5 x 450 - 0.338 x 450) = 57.57 kN-m
Mux1 = Muc + Mus(Total)
= 57.57 + (42.54)
= 100.11 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1206 0.00285 352.57 8.92 343.65 414.57 102 42.29
2 1206 0.00008 15.78 0.69 15.09 18.21 -102 -1.86
Total432.78 40.43
xuy = 258 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 258/300 = 0.309
Puc = 0.309 x 20.00 x 450 x 300
= 835.31 kN
Puy1 = Puc + Pus(Total)
= 835.31 + (432.78)
= 1268.09 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 258/300
= 0.358
Muc = 835.31 x (0.5 x 300 - 0.358 x 300) = 35.71 kN-m
Muy1 = Muc + Mus(Total)
= 35.71 + (40.43)
= 76.14 kN-m
Pu/Puz = 0.647
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.647, an = 1.746
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((91.36/100.11)1.746) + ((25.17/76.14)1.746)
= 0.997 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG2 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 12 nos. (2413 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 1639.13 kN
MomentX,(Mx) = 66.74 kN-m
MomentY,(My) = 0.99 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C100.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 35.651 kN.m
My_MinEccen = 32.783 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 66.735 kN.m
My = max(My,My_MinEccen) + MuaddY = 32.783 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 66.74/1639.13 = 41 mm
Actual eccenY = My / P = 0.99/1639.13 = 1 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(41,22) = 41 mm
eccenY = max(Actual eccenY,eccenYMin) = max(1,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(41 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 16 nos. (3217 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00309 354.70 8.92 345.78 139.05 177 24.61
2 402 0.00266 349.26 8.92 340.34 136.86 126 17.30
3 402 0.00223 336.22 8.92 327.30 131.62 76 9.98
4 402 0.00180 317.35 8.83 308.52 124.06 25 3.14
5 402 0.00137 274.06 8.04 266.03 106.98 -25 -2.70
6 402 0.00094 188.00 6.41 181.59 73.02 -76 -5.54
7 402 0.00051 101.94 3.97 97.97 39.40 -126 -4.98
8 402 0.00008 15.87 0.69 15.18 6.10 -177 -1.08
Total757.08 40.73
xux = 411 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 411/450 = 0.329
Puc = 0.329 x 20.00 x 300 x 450
= 888.47 kN
Pux1 = Puc + Pus(Total)
= 888.47 + (757.08)
= 1645.55 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 411/450
= 0.380
Muc = 888.47 x (0.5 x 450 - 0.380 x 450) = 47.88 kN-m
Mux1 = Muc + Mus(Total)
= 47.88 + (40.73)
= 88.61 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1608 0.00292 353.22 8.92 344.30 553.80 102 56.49
2 1608 0.00047 93.03 3.67 89.37 143.74 -102 -14.66
Total697.54 41.83
xuy = 291 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 291/300 = 0.349
Puc = 0.349 x 20.00 x 450 x 300
= 941.63 kN
Puy1 = Puc + Pus(Total)
= 941.63 + (697.54)
= 1639.17 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 291/300
= 0.403
Muc = 941.63 x (0.5 x 300 - 0.403 x 300) = 27.40 kN-m
Muy1 = Muc + Mus(Total)
= 27.40 + (41.83)
= 69.23 kN-m
Pu/Puz = 0.749
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.749, an = 1.916
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((66.74/88.61)1.916) + ((32.78/69.23)1.916)
= 0.820 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG2 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 16 nos. (3217 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG3 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 137.99 kN
MomentX,(Mx) = 49.09 kN-m
MomentY,(My) = 26.54 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C110.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 2.856 kN.m
My_MinEccen = 2.760 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 49.086 kN.m
My = max(My,My_MinEccen) + MuaddY = 26.543 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 49.09/137.99 = 356 mm
Actual eccenY = My / P = 26.54/137.99 = 192 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(356,21) = 356 mm
eccenY = max(Actual eccenY,eccenYMin) = max(192,20) = 192 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(356 mm) > 0.05 x 300(15 mm)
and eccenY(192 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 6 nos. (1206 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00225 336.80 8.92 327.88 131.85 177 23.34
2 402 -0.00238 -341.53 0.00 -341.53 -137.34 0 0.00
3 402 -0.00700 -360.90 0.00 -360.90 -145.13 -177 25.69
Total-150.62 49.02
xux = 134 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 134/450 = 0.107
Puc = 0.107 x 20.00 x 300 x 450
= 289.51 kN
Pux1 = Puc + Pus(Total)
= 289.51 + (-150.62)
= 138.90 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 134/450
= 0.124
Muc = 289.51 x (0.5 x 450 - 0.124 x 450) = 49.00 kN-m
Mux1 = Muc + Mus(Total)
= 49.00 + (49.02)
= 98.02 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 603 0.00111 221.14 7.14 214.00 129.08 102 13.17
2 603 -0.00907 -360.90 0.00 -360.90 -217.69 -102 22.20
Total-88.61 35.37
xuy = 70 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 70/300 = 0.084
Puc = 0.084 x 20.00 x 450 x 300
= 227.34 kN
Puy1 = Puc + Pus(Total)
= 227.34 + (-88.61)
= 138.73 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 70/300
= 0.097
Muc = 227.34 x (0.5 x 300 - 0.097 x 300) = 27.46 kN-m
Muy1 = Muc + Mus(Total)
= 27.46 + (35.37)
= 62.84 kN-m
Pu/Puz = 0.087
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.087, an = 1.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((49.09/98.02)1.000) + ((26.54/62.84)1.000)
= 0.923 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG3 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 6 nos. (1206 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 323.16 kN
MomentX,(Mx) = 45.84 kN-m
MomentY,(My) = 23.18 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C110.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 6.689 kN.m
My_MinEccen = 6.463 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 45.835 kN.m
My = max(My,My_MinEccen) + MuaddY = 23.180 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 45.84/323.16 = 142 mm
Actual eccenY = My / P = 23.18/323.16 = 72 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(142,21) = 142 mm
eccenY = max(Actual eccenY,eccenYMin) = max(72,20) = 72 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(142 mm) > 0.05 x 300(15 mm)
and eccenY(72 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00263 348.50 8.92 339.58 76.81 179 13.75
2 226 0.00094 188.54 6.43 182.11 41.19 90 3.69
3 226 -0.00075 -149.29 0.00 -149.29 -33.77 0 0.00
4 226 -0.00244 -343.46 0.00 -343.46 -77.69 -90 6.95
5 226 -0.00412 -360.90 0.00 -360.90 -81.63 -179 14.61
Total-75.09 39.00
xux = 185 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 185/450 = 0.148
Puc = 0.148 x 20.00 x 300 x 450
= 400.57 kN
Pux1 = Puc + Pus(Total)
= 400.57 + (-75.09)
= 325.48 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 185/450
= 0.171
Muc = 400.57 x (0.5 x 450 - 0.171 x 450) = 59.23 kN-m
Mux1 = Muc + Mus(Total)
= 59.23 + (39.00)
= 98.23 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00201 327.98 8.92 319.06 180.43 104 18.76
2 565 -0.00475 -360.90 0.00 -360.90 -204.08 -104 21.22
Total-23.66 39.99
xuy = 108 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 108/300 = 0.129
Puc = 0.129 x 20.00 x 450 x 300
= 349.31 kN
Puy1 = Puc + Pus(Total)
= 349.31 + (-23.66)
= 325.65 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 108/300
= 0.149
Muc = 349.31 x (0.5 x 300 - 0.149 x 300) = 36.73 kN-m
Muy1 = Muc + Mus(Total)
= 36.73 + (39.99)
= 76.72 kN-m
Pu/Puz = 0.208
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.208, an = 1.013
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((45.84/98.23)1.013) + ((23.18/76.72)1.013)
= 0.760 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG3 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 491.14 kN
MomentX,(Mx) = 38.31 kN-m
MomentY,(My) = 19.18 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C110.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 10.167 kN.m
My_MinEccen = 9.823 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 38.309 kN.m
My = max(My,My_MinEccen) + MuaddY = 19.176 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 38.31/491.14 = 78 mm
Actual eccenY = My / P = 19.18/491.14 = 39 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(78,21) = 78 mm
eccenY = max(Actual eccenY,eccenYMin) = max(39,20) = 39 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(78 mm) > 0.05 x 300(15 mm)
and eccenY(39 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00279 352.09 8.92 343.17 77.62 179 13.89
2 226 0.00142 283.33 8.16 275.17 62.24 90 5.57
3 226 0.00004 8.11 0.36 7.75 1.75 0 0.00
4 226 -0.00134 -267.11 0.00 -267.11 -60.42 -90 5.41
5 226 -0.00271 -350.56 0.00 -350.56 -79.29 -179 14.19
Total1.90 39.07
xux = 228 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 228/450 = 0.182
Puc = 0.182 x 20.00 x 300 x 450
= 491.70 kN
Pux1 = Puc + Pus(Total)
= 491.70 + (1.90)
= 493.60 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 228/450
= 0.210
Muc = 491.70 x (0.5 x 450 - 0.210 x 450) = 64.07 kN-m
Mux1 = Muc + Mus(Total)
= 64.07 + (39.07)
= 103.14 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00245 343.86 8.92 334.94 189.40 104 19.70
2 565 -0.00229 -338.43 0.00 -338.43 -191.38 -104 19.90
Total-1.97 39.60
xuy = 154 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 154/300 = 0.184
Puc = 0.184 x 20.00 x 450 x 300
= 497.39 kN
Puy1 = Puc + Pus(Total)
= 497.39 + (-1.97)
= 495.42 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 154/300
= 0.213
Muc = 497.39 x (0.5 x 300 - 0.213 x 300) = 42.84 kN-m
Muy1 = Muc + Mus(Total)
= 42.84 + (39.60)
= 82.45 kN-m
Pu/Puz = 0.315
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.315, an = 1.192
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((38.31/103.14)1.192) + ((19.18/82.45)1.192)
= 0.483 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG3 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 740.97 kN
MomentX,(Mx) = 27.01 kN-m
MomentY,(My) = 13.81 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C110.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 15.338 kN.m
My_MinEccen = 14.819 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 27.009 kN.m
My = max(My,My_MinEccen) + MuaddY = 14.819 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 27.01/740.97 = 36 mm
Actual eccenY = My / P = 13.81/740.97 = 19 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(36,21) = 36 mm
eccenY = max(Actual eccenY,eccenYMin) = max(19,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(36 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00295 353.48 8.92 344.56 77.94 179 13.95
2 226 0.00188 322.58 8.89 313.69 70.96 90 6.35
3 226 0.00082 163.47 5.80 157.67 35.66 0 0.00
4 226 -0.00025 -49.95 0.00 -49.95 -11.30 -90 1.01
5 226 -0.00132 -263.36 0.00 -263.36 -59.57 -179 10.66
Total113.69 31.98
xux = 294 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 294/450 = 0.235
Puc = 0.235 x 20.00 x 300 x 450
= 634.08 kN
Pux1 = Puc + Pus(Total)
= 634.08 + (113.69)
= 747.77 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 294/450
= 0.271
Muc = 634.08 x (0.5 x 450 - 0.271 x 450) = 65.23 kN-m
Mux1 = Muc + Mus(Total)
= 65.23 + (31.98)
= 97.21 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00270 350.29 8.92 341.37 193.04 104 20.08
2 565 -0.00091 -182.11 0.00 -182.11 -102.98 -104 10.71
Total90.06 30.79
xuy = 202 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 202/300 = 0.242
Puc = 0.242 x 20.00 x 450 x 300
= 653.06 kN
Puy1 = Puc + Pus(Total)
= 653.06 + (90.06)
= 743.12 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 202/300
= 0.280
Muc = 653.06 x (0.5 x 300 - 0.280 x 300) = 43.20 kN-m
Muy1 = Muc + Mus(Total)
= 43.20 + (30.79)
= 73.99 kN-m
Pu/Puz = 0.476
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.476, an = 1.460
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((27.01/97.21)1.460) + ((14.82/73.99)1.460)
= 0.250 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG3 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 854.17 kN
MomentX,(Mx) = 15.00 kN-m
MomentY,(My) = 6.47 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C110.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 18.578 kN.m
My_MinEccen = 17.083 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 18.578 kN.m
My = max(My,My_MinEccen) + MuaddY = 17.083 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 15.00/854.17 = 18 mm
Actual eccenY = My / P = 6.47/854.17 = 8 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(18,22) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(8,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00300 353.92 8.92 345.00 78.04 179 13.97
2 226 0.00203 328.98 8.92 320.06 72.40 90 6.48
3 226 0.00107 213.04 6.97 206.07 46.61 0 0.00
4 226 0.00010 19.34 0.84 18.50 4.18 -90 -0.37
5 226 -0.00087 -174.36 0.00 -174.36 -39.44 -179 7.06
Total161.79 27.13
xux = 323 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 323/450 = 0.259
Puc = 0.259 x 20.00 x 300 x 450
= 698.63 kN
Pux1 = Puc + Pus(Total)
= 698.63 + (161.79)
= 860.42 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 323/450
= 0.299
Muc = 698.63 x (0.5 x 450 - 0.299 x 450) = 63.19 kN-m
Mux1 = Muc + Mus(Total)
= 63.19 + (27.13)
= 90.32 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00278 351.95 8.92 343.03 193.98 104 20.17
2 565 -0.00049 -98.54 0.00 -98.54 -55.72 -104 5.80
Total138.25 25.97
xuy = 223 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 223/300 = 0.267
Puc = 0.267 x 20.00 x 450 x 300
= 721.41 kN
Puy1 = Puc + Pus(Total)
= 721.41 + (138.25)
= 859.66 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 223/300
= 0.309
Muc = 721.41 x (0.5 x 300 - 0.309 x 300) = 41.39 kN-m
Muy1 = Muc + Mus(Total)
= 41.39 + (25.97)
= 67.36 kN-m
Pu/Puz = 0.549
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.549, an = 1.581
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((18.58/90.32)1.581) + ((17.08/67.36)1.581)
= 0.196 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG3 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG4 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3450 mm
From analysis results,loads on column
Axial load,(P) = 286.83 kN
MomentX,(Mx) = 62.42 kN-m
MomentY,(My) = 63.98 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C120.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3450 x 1.000 ) / 450 = 7.67 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3450 x 1.000 ) / 300 = 11.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 6.282 kN.m
My_MinEccen = 5.737 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 62.421 kN.m
My = max(My,My_MinEccen) + MuaddY = 63.978 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 62.42/286.83 = 218 mm
Actual eccenY = My / P = 63.98/286.83 = 223 mm
eccenXMin = (L/500) + (D/30) = (3450/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3450/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(218,22) = 218 mm
eccenY = max(Actual eccenY,eccenYMin) = max(223,20) = 223 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(218 mm) > 0.05 x 300(15 mm)
and eccenY(223 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 12 nos. (2413 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00264 348.59 8.92 339.67 136.59 177 24.18
2 402 0.00136 271.87 8.00 263.86 106.10 106 11.27
3 402 0.00008 16.72 0.73 15.99 6.43 35 0.23
4 402 -0.00119 -238.43 0.00 -238.43 -95.88 -35 3.39
5 402 -0.00247 -344.29 0.00 -344.29 -138.45 -106 14.70
6 402 -0.00374 -360.41 0.00 -360.41 -144.93 -177 25.65
Total-130.13 79.42
xux = 194 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 194/450 = 0.155
Puc = 0.155 x 20.00 x 300 x 450
= 419.55 kN
Pux1 = Puc + Pus(Total)
= 419.55 + (-130.13)
= 289.42 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 194/450
= 0.180
Muc = 419.55 x (0.5 x 450 - 0.180 x 450) = 60.50 kN-m
Mux1 = Muc + Mus(Total)
= 60.50 + (79.42)
= 139.92 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1206 0.00192 324.54 8.90 315.64 380.78 102 38.84
2 1206 -0.00482 -360.90 0.00 -360.90 -435.38 -102 44.41
Total-54.60 83.25
xuy = 106 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 106/300 = 0.127
Puc = 0.127 x 20.00 x 450 x 300
= 343.62 kN
Puy1 = Puc + Pus(Total)
= 343.62 + (-54.60)
= 289.02 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 106/300
= 0.147
Muc = 343.62 x (0.5 x 300 - 0.147 x 300) = 36.38 kN-m
Muy1 = Muc + Mus(Total)
= 36.38 + (83.25)
= 119.63 kN-m
Pu/Puz = 0.148
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.148, an = 1.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((62.42/139.92)1.000) + ((63.98/119.63)1.000)
= 0.981 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG4 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 12 nos. (2413 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3450 mm
From analysis results,loads on column
Axial load,(P) = 594.33 kN
MomentX,(Mx) = 42.37 kN-m
MomentY,(My) = 56.23 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C120.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3450 x 1.000 ) / 450 = 7.67 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3450 x 1.000 ) / 300 = 11.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 13.016 kN.m
My_MinEccen = 11.887 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 42.366 kN.m
My = max(My,My_MinEccen) + MuaddY = 56.229 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 42.37/594.33 = 71 mm
Actual eccenY = My / P = 56.23/594.33 = 95 mm
eccenXMin = (L/500) + (D/30) = (3450/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3450/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(71,22) = 71 mm
eccenY = max(Actual eccenY,eccenYMin) = max(95,20) = 95 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(71 mm) > 0.05 x 300(15 mm)
and eccenY(95 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00286 352.71 8.92 343.79 77.76 179 13.92
2 226 0.00163 306.36 8.61 297.75 67.35 90 6.03
3 226 0.00039 77.78 3.13 74.65 16.88 0 0.00
4 226 -0.00085 -169.73 0.00 -169.73 -38.39 -90 3.44
5 226 -0.00209 -330.90 0.00 -330.90 -74.85 -179 13.40
Total48.76 36.78
xux = 253 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 253/450 = 0.203
Puc = 0.203 x 20.00 x 300 x 450
= 546.75 kN
Pux1 = Puc + Pus(Total)
= 546.75 + (48.76)
= 595.51 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 253/450
= 0.234
Muc = 546.75 x (0.5 x 450 - 0.234 x 450) = 65.45 kN-m
Mux1 = Muc + Mus(Total)
= 65.45 + (36.78)
= 102.23 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00259 347.43 8.92 338.51 191.42 104 19.91
2 565 -0.00152 -296.65 0.00 -296.65 -167.75 -104 17.45
Total23.67 37.35
xuy = 177 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 177/300 = 0.212
Puc = 0.212 x 20.00 x 450 x 300
= 573.33 kN
Puy1 = Puc + Pus(Total)
= 573.33 + (23.67)
= 597.00 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 177/300
= 0.245
Muc = 573.33 x (0.5 x 300 - 0.245 x 300) = 43.80 kN-m
Muy1 = Muc + Mus(Total)
= 43.80 + (37.35)
= 81.15 kN-m
Pu/Puz = 0.382
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.382, an = 1.303
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((42.37/102.23)1.303) + ((56.23/81.15)1.303)
= 0.937 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG4 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3450 mm
From analysis results,loads on column
Axial load,(P) = 894.06 kN
MomentX,(Mx) = 55.58 kN-m
MomentY,(My) = 55.13 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C120.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3450 x 1.000 ) / 450 = 7.67 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3450 x 1.000 ) / 300 = 11.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 19.580 kN.m
My_MinEccen = 17.881 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 55.578 kN.m
My = max(My,My_MinEccen) + MuaddY = 55.133 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 55.58/894.06 = 62 mm
Actual eccenY = My / P = 55.13/894.06 = 62 mm
eccenXMin = (L/500) + (D/30) = (3450/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3450/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(62,22) = 62 mm
eccenY = max(Actual eccenY,eccenYMin) = max(62,20) = 62 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(62 mm) > 0.05 x 300(15 mm)
and eccenY(62 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 6 nos. + #16 - 4 nos. (1483 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00298 353.75 8.92 344.83 138.67 177 24.54
2 226 0.00202 328.48 8.92 319.56 72.28 88 6.32
3 226 0.00108 215.68 7.03 208.65 47.20 0 0.00
4 226 0.00014 27.33 1.18 26.15 5.92 -88 -0.52
5 402 -0.00083 -165.33 0.00 -165.33 -66.48 -177 11.77
Total197.58 42.12
xux = 325 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 325/450 = 0.260
Puc = 0.260 x 20.00 x 300 x 450
= 702.42 kN
Pux1 = Puc + Pus(Total)
= 702.42 + (197.58)
= 900.00 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 325/450
= 0.301
Muc = 702.42 x (0.5 x 450 - 0.301 x 450) = 63.02 kN-m
Mux1 = Muc + Mus(Total)
= 63.02 + (42.12)
= 105.14 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 741 0.00274 351.32 8.92 342.40 253.86 102 25.89
2 741 -0.00048 -96.44 0.00 -96.44 -71.51 -102 7.29
Total182.36 33.19
xuy = 221 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 221/300 = 0.266
Puc = 0.266 x 20.00 x 450 x 300
= 717.61 kN
Puy1 = Puc + Pus(Total)
= 717.61 + (182.36)
= 899.97 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 221/300
= 0.307
Muc = 717.61 x (0.5 x 300 - 0.307 x 300) = 41.52 kN-m
Muy1 = Muc + Mus(Total)
= 41.52 + (33.19)
= 74.71 kN-m
Pu/Puz = 0.538
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.538, an = 1.563
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((55.58/105.14)1.563) + ((55.13/74.71)1.563)
= 0.991 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG4 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 6 nos. + #16 - 4 nos. (1483 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3450 mm
From analysis results,loads on column
Axial load,(P) = 1289.95 kN
MomentX,(Mx) = 51.02 kN-m
MomentY,(My) = 56.42 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C120.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3450 x 1.000 ) / 450 = 7.67 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3450 x 1.000 ) / 300 = 11.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 28.250 kN.m
My_MinEccen = 25.799 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 51.019 kN.m
My = max(My,My_MinEccen) + MuaddY = 56.421 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 51.02/1289.95 = 40 mm
Actual eccenY = My / P = 56.42/1289.95 = 44 mm
eccenXMin = (L/500) + (D/30) = (3450/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3450/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(40,22) = 40 mm
eccenY = max(Actual eccenY,eccenYMin) = max(44,20) = 44 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(40 mm) > 0.05 x 300(15 mm)
and eccenY(44 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 12 nos. (2413 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00305 354.35 8.92 345.43 138.91 177 24.59
2 402 0.00239 342.05 8.92 333.13 133.96 106 14.23
3 402 0.00173 312.79 8.75 304.04 122.26 35 4.33
4 402 0.00107 213.16 6.97 206.19 82.91 -35 -2.94
5 402 0.00040 80.79 3.24 77.55 31.19 -106 -3.31
6 402 -0.00026 -51.57 0.00 -51.57 -20.74 -177 3.67
Total488.48 40.56
xux = 374 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 374/450 = 0.300
Puc = 0.300 x 20.00 x 300 x 450
= 808.73 kN
Pux1 = Puc + Pus(Total)
= 808.73 + (488.48)
= 1297.22 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 374/450
= 0.346
Muc = 808.73 x (0.5 x 450 - 0.346 x 450) = 56.00 kN-m
Mux1 = Muc + Mus(Total)
= 56.00 + (40.56)
= 96.56 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1206 0.00286 352.67 8.92 343.75 414.70 102 42.30
2 1206 0.00014 28.00 1.21 26.79 32.32 -102 -3.30
Total447.02 39.00
xuy = 263 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 263/300 = 0.315
Puc = 0.315 x 20.00 x 450 x 300
= 850.50 kN
Puy1 = Puc + Pus(Total)
= 850.50 + (447.02)
= 1297.52 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 263/300
= 0.364
Muc = 850.50 x (0.5 x 300 - 0.364 x 300) = 34.70 kN-m
Muy1 = Muc + Mus(Total)
= 34.70 + (39.00)
= 73.70 kN-m
Pu/Puz = 0.663
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.663, an = 1.772
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((51.02/96.56)1.772) + ((56.42/73.70)1.772)
= 0.946 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG4 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 12 nos. (2413 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3675 mm
From analysis results,loads on column
Axial load,(P) = 1564.72 kN
MomentX,(Mx) = 19.13 kN-m
MomentY,(My) = 39.26 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C120.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3675 x 1.000 ) / 450 = 8.17 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3675 x 1.000 ) / 300 = 12.25 > 12.00,
column is slender in this direction.
Puz = (0.45 x fck x Ag) + (0.75 x fy - 0.45 x fck) x Ast
= 0.45 x 20.00 x 135000.01 + (0.75 x 415.00 - 0.45 x 20.00 ) x 2814.87 = 2065793.701 N
k1 = 0.193600
k2 = 0.032400 MPa
Pub = (k1 x fck x Ag) + (k2 x Ast x 100)
= (0.193600 x 20.00 x 135000.01) + (0.03 x 2814.87 x 100) = 531840.210 N
k = (Puz - Pu) / (Puz - Pub) <= 1
= (2065793.701 - 1564720.581) / (2065793.701 - 531840.210) = 0.326655
MuaddY = Pu x (Width/2000) x (Lo/Width) x (Lo/Width) x k
= 1564720.58 x (300.00/2000) x (3675.00/300.00) x (3675.00/300.00) x 0.326655
= 11505067.825 N.mm = 11.505 kN.m
Mx_MinEccen = 34.972 kN.m
My_MinEccen = 31.294 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 34.972 kN.m
My = max(My,My_MinEccen) + MuaddY = 50.764 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 19.13/1564.72 = 12 mm
Actual eccenY = My / P = 39.26/1564.72 = 25 mm
eccenXMin = (L/500) + (D/30) = (3675/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3675/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(12,22) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(25,20) = 25 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(25 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 14 nos. (2815 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00310 354.76 8.92 345.84 139.07 177 24.62
2 402 0.00260 347.81 8.92 338.89 136.28 118 16.08
3 402 0.00211 331.83 8.92 322.91 129.85 59 7.66
4 402 0.00162 305.53 8.59 296.94 119.40 0 0.00
5 402 0.00112 224.81 7.21 217.60 87.50 -59 -5.16
6 402 0.00063 126.09 4.74 121.35 48.80 -118 -5.76
7 402 0.00014 27.37 1.18 26.19 10.53 -177 -1.86
Total671.43 35.57
xux = 418 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 418/450 = 0.335
Puc = 0.335 x 20.00 x 300 x 450
= 903.66 kN
Pux1 = Puc + Pus(Total)
= 903.66 + (671.43)
= 1575.09 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 418/450
= 0.387
Muc = 903.66 x (0.5 x 450 - 0.387 x 450) = 46.05 kN-m
Mux1 = Muc + Mus(Total)
= 46.05 + (35.57)
= 81.62 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1407 0.00293 353.26 8.92 344.34 484.63 102 49.43
2 1407 0.00049 97.89 3.83 94.06 132.38 -102 -13.50
Total617.01 35.93
xuy = 293 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 293/300 = 0.352
Puc = 0.352 x 20.00 x 450 x 300
= 949.22 kN
Puy1 = Puc + Pus(Total)
= 949.22 + (617.01)
= 1566.23 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 293/300
= 0.406
Muc = 949.22 x (0.5 x 300 - 0.406 x 300) = 26.70 kN-m
Muy1 = Muc + Mus(Total)
= 26.70 + (35.93)
= 62.63 kN-m
Pu/Puz = 0.757
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.757, an = 1.929
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((34.97/81.62)1.929) + ((50.76/62.63)1.929)
= 0.862 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG4 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 14 nos. (2815 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG5 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 296.80 kN
MomentX,(Mx) = 41.32 kN-m
MomentY,(My) = 64.49 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C130.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 6.144 kN.m
My_MinEccen = 5.936 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 41.317 kN.m
My = max(My,My_MinEccen) + MuaddY = 64.493 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 41.32/296.80 = 139 mm
Actual eccenY = My / P = 64.49/296.80 = 217 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(139,21) = 139 mm
eccenY = max(Actual eccenY,eccenYMin) = max(217,20) = 217 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(139 mm) > 0.05 x 300(15 mm)
and eccenY(217 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 16 nos. (1810 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00266 349.12 8.92 340.20 76.95 179 13.77
2 226 0.00172 312.15 8.74 303.41 68.63 128 8.77
3 226 0.00078 155.75 5.59 150.16 33.97 77 2.61
4 226 -0.00016 -31.95 0.00 -31.95 -7.23 26 -0.18
5 226 -0.00110 -219.66 0.00 -219.66 -49.69 -26 1.27
6 226 -0.00204 -329.09 0.00 -329.09 -74.44 -77 5.71
7 226 -0.00298 -353.68 0.00 -353.68 -80.00 -128 10.23
8 226 -0.00391 -360.90 0.00 -360.90 -81.63 -179 14.61
Total-113.44 56.79
xux = 191 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 191/450 = 0.153
Puc = 0.153 x 20.00 x 300 x 450
= 411.96 kN
Pux1 = Puc + Pus(Total)
= 411.96 + (-113.44)
= 298.52 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 191/450
= 0.176
Muc = 411.96 x (0.5 x 450 - 0.176 x 450) = 60.01 kN-m
Mux1 = Muc + Mus(Total)
= 60.01 + (56.79)
= 116.80 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 905 0.00196 326.13 8.92 317.22 287.01 104 29.85
2 905 -0.00502 -360.90 0.00 -360.90 -326.53 -104 33.96
Total-39.52 63.81
xuy = 104 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 104/300 = 0.125
Puc = 0.125 x 20.00 x 450 x 300
= 337.92 kN
Puy1 = Puc + Pus(Total)
= 337.92 + (-39.52)
= 298.40 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 104/300
= 0.145
Muc = 337.92 x (0.5 x 300 - 0.145 x 300) = 36.03 kN-m
Muy1 = Muc + Mus(Total)
= 36.03 + (63.81)
= 99.84 kN-m
Pu/Puz = 0.168
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.168, an = 1.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((41.32/116.80)1.000) + ((64.49/99.84)1.000)
= 1.000 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG5 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 16 nos. (1810 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 611.13 kN
MomentX,(Mx) = 35.92 kN-m
MomentY,(My) = 57.31 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C130.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 12.650 kN.m
My_MinEccen = 12.223 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 35.918 kN.m
My = max(My,My_MinEccen) + MuaddY = 57.310 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 35.92/611.13 = 59 mm
Actual eccenY = My / P = 57.31/611.13 = 94 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(59,21) = 59 mm
eccenY = max(Actual eccenY,eccenYMin) = max(94,20) = 94 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(59 mm) > 0.05 x 300(15 mm)
and eccenY(94 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00288 352.82 8.92 343.90 77.79 179 13.92
2 226 0.00166 308.86 8.67 300.19 67.90 90 6.08
3 226 0.00045 90.48 3.58 86.90 19.66 0 0.00
4 226 -0.00076 -151.98 0.00 -151.98 -34.38 -90 3.08
5 226 -0.00197 -326.72 0.00 -326.72 -73.90 -179 13.23
Total57.07 36.31
xux = 258 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 258/450 = 0.207
Puc = 0.207 x 20.00 x 300 x 450
= 558.14 kN
Pux1 = Puc + Pus(Total)
= 558.14 + (57.07)
= 615.21 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 258/450
= 0.239
Muc = 558.14 x (0.5 x 450 - 0.239 x 450) = 65.59 kN-m
Mux1 = Muc + Mus(Total)
= 65.59 + (36.31)
= 101.89 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00261 347.89 8.92 338.97 191.68 104 19.93
2 565 -0.00143 -285.21 0.00 -285.21 -161.28 -104 16.77
Total30.40 36.71
xuy = 180 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 180/300 = 0.217
Puc = 0.217 x 20.00 x 450 x 300
= 584.72 kN
Puy1 = Puc + Pus(Total)
= 584.72 + (30.40)
= 615.12 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 180/300
= 0.250
Muc = 584.72 x (0.5 x 300 - 0.250 x 300) = 43.81 kN-m
Muy1 = Muc + Mus(Total)
= 43.81 + (36.71)
= 80.52 kN-m
Pu/Puz = 0.393
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.393, an = 1.321
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((35.92/101.89)1.321) + ((57.31/80.52)1.321)
= 0.890 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG5 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 919.74 kN
MomentX,(Mx) = 32.03 kN-m
MomentY,(My) = 56.33 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C130.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 19.039 kN.m
My_MinEccen = 18.395 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 32.026 kN.m
My = max(My,My_MinEccen) + MuaddY = 56.329 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 32.03/919.74 = 35 mm
Actual eccenY = My / P = 56.33/919.74 = 61 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(35,21) = 35 mm
eccenY = max(Actual eccenY,eccenYMin) = max(61,20) = 61 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(35 mm) > 0.05 x 300(15 mm)
and eccenY(61 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 6 nos. (1206 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00301 353.99 8.92 345.07 138.76 177 24.56
2 402 0.00120 240.51 7.50 233.01 93.70 0 0.00
3 402 -0.00060 -120.95 0.00 -120.95 -48.64 -177 8.61
Total183.82 33.17
xux = 343 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 343/450 = 0.274
Puc = 0.274 x 20.00 x 300 x 450
= 740.39 kN
Pux1 = Puc + Pus(Total)
= 740.39 + (183.82)
= 924.21 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 343/450
= 0.317
Muc = 740.39 x (0.5 x 450 - 0.317 x 450) = 61.01 kN-m
Mux1 = Muc + Mus(Total)
= 61.01 + (33.17)
= 94.18 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 603 0.00278 351.94 8.92 343.02 206.90 102 21.10
2 603 -0.00030 -60.24 0.00 -60.24 -36.34 -102 3.71
Total170.57 24.81
xuy = 232 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 232/300 = 0.278
Puc = 0.278 x 20.00 x 450 x 300
= 751.78 kN
Puy1 = Puc + Pus(Total)
= 751.78 + (170.57)
= 922.35 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 232/300
= 0.322
Muc = 751.78 x (0.5 x 300 - 0.322 x 300) = 40.20 kN-m
Muy1 = Muc + Mus(Total)
= 40.20 + (24.81)
= 65.01 kN-m
Pu/Puz = 0.582
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.582, an = 1.637
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((32.03/94.18)1.637) + ((56.33/65.01)1.637)
= 0.962 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG5 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 6 nos. (1206 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1216.66 kN
MomentX,(Mx) = 26.15 kN-m
MomentY,(My) = 56.89 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C130.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 25.185 kN.m
My_MinEccen = 24.333 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 26.152 kN.m
My = max(My,My_MinEccen) + MuaddY = 56.888 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 26.15/1216.66 = 21 mm
Actual eccenY = My / P = 56.89/1216.66 = 47 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(21,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(47,20) = 47 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(47 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 6 nos. + #16 - 6 nos. (1885 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00307 354.47 8.92 345.55 138.96 177 24.60
2 226 0.00242 343.09 8.92 334.17 75.59 106 8.00
3 226 0.00179 316.99 8.83 308.16 69.71 37 2.55
4 402 0.00115 230.10 7.31 222.79 89.59 -35 -3.10
5 226 0.00051 101.22 3.94 97.28 22.00 -106 -2.33
6 402 -0.00014 -27.66 0.00 -27.66 -11.12 -177 1.97
Total384.72 31.68
xux = 387 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 387/450 = 0.309
Puc = 0.309 x 20.00 x 300 x 450
= 835.31 kN
Pux1 = Puc + Pus(Total)
= 835.31 + (384.72)
= 1220.03 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 387/450
= 0.358
Muc = 835.31 x (0.5 x 450 - 0.358 x 450) = 53.56 kN-m
Mux1 = Muc + Mus(Total)
= 53.56 + (31.68)
= 85.25 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 942 0.00287 352.77 8.92 343.85 324.07 102 33.06
2 942 0.00020 39.79 1.69 38.10 35.91 -102 -3.66
Total359.99 29.39
xuy = 267 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 267/300 = 0.321
Puc = 0.321 x 20.00 x 450 x 300
= 865.69 kN
Puy1 = Puc + Pus(Total)
= 865.69 + (359.99)
= 1225.67 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 267/300
= 0.370
Muc = 865.69 x (0.5 x 300 - 0.370 x 300) = 33.63 kN-m
Muy1 = Muc + Mus(Total)
= 33.63 + (29.39)
= 63.02 kN-m
Pu/Puz = 0.682
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.682, an = 1.803
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((26.15/85.25)1.803) + ((56.89/63.02)1.803)
= 0.950 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG5 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 6 nos. + #16 - 6 nos. (1885 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 1503.07 kN
MomentX,(Mx) = 13.88 kN-m
MomentY,(My) = 38.95 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C130.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 32.692 kN.m
My_MinEccen = 30.062 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 32.692 kN.m
My = max(My,My_MinEccen) + MuaddY = 38.949 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 13.88/1503.07 = 9 mm
Actual eccenY = My / P = 38.95/1503.07 = 26 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(9,22) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(26,20) = 26 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(26 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 12 nos. + #16 - 4 nos. (2161 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00313 355.01 8.92 346.09 139.17 177 24.63
2 226 0.00272 350.83 8.92 341.91 77.34 125 9.67
3 226 0.00233 339.98 8.92 331.06 74.88 75 5.62
4 226 0.00194 325.70 8.91 316.78 71.66 25 1.79
5 226 0.00156 299.65 8.48 291.17 65.86 -25 -1.65
6 226 0.00117 233.33 7.37 225.96 51.11 -75 -3.83
7 226 0.00078 155.56 5.59 149.97 33.92 -125 -4.24
8 402 0.00037 74.67 3.02 71.65 28.81 -177 -5.10
Total542.75 26.89
xux = 450 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 450/450 = 0.360
Puc = 0.360 x 20.00 x 300 x 450
= 972.00 kN
Pux1 = Puc + Pus(Total)
= 972.00 + (542.75)
= 1514.75 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 450/450
= 0.416
Muc = 972.00 x (0.5 x 450 - 0.416 x 450) = 36.74 kN-m
Mux1 = Muc + Mus(Total)
= 36.74 + (26.89)
= 63.63 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1081 0.00288 352.85 8.92 343.93 371.69 102 37.91
2 1081 0.00065 130.43 4.87 125.56 135.69 -102 -13.84
Total507.38 24.07
xuy = 312 mm               Puc = C1.fck.B.D
ku = 1.039
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.001
C1 = 0.446 x (1 - C3/6) = 0.372
Puc = 0.372 x 20.00 x 450 x 300
= 1003.24 kN
Puy1 = Puc + Pus(Total)
= 1003.24 + (507.38)
= 1510.62 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.428
Muc = 1003.24 x (0.5 x 300 - 0.428 x 300) = 21.53 kN-m
Muy1 = Muc + Mus(Total)
= 21.53 + (24.07)
= 45.60 kN-m
Pu/Puz = 0.805
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.805, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((32.69/63.63)2.000) + ((38.95/45.60)2.000)
= 0.993 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG5 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 12 nos. + #16 - 4 nos. (2161 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG6 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 278.42 kN
MomentX,(Mx) = 58.86 kN-m
MomentY,(My) = 15.28 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C140.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 5.763 kN.m
My_MinEccen = 5.568 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 58.861 kN.m
My = max(My,My_MinEccen) + MuaddY = 15.282 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 58.86/278.42 = 211 mm
Actual eccenY = My / P = 15.28/278.42 = 55 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(211,21) = 211 mm
eccenY = max(Actual eccenY,eccenYMin) = max(55,20) = 55 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(211 mm) > 0.05 x 300(15 mm)
and eccenY(55 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00257 347.04 8.92 338.12 76.48 179 13.69
2 226 0.00077 154.96 5.57 149.39 33.79 90 3.02
3 226 -0.00103 -205.05 0.00 -205.05 -46.38 0 0.00
4 226 -0.00283 -352.37 0.00 -352.37 -79.70 -90 7.13
5 226 -0.00463 -360.90 0.00 -360.90 -81.63 -179 14.61
Total-97.45 38.46
xux = 174 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 174/450 = 0.139
Puc = 0.139 x 20.00 x 300 x 450
= 375.89 kN
Pux1 = Puc + Pus(Total)
= 375.89 + (-97.45)
= 278.44 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 174/450
= 0.161
Muc = 375.89 x (0.5 x 450 - 0.161 x 450) = 57.36 kN-m
Mux1 = Muc + Mus(Total)
= 57.36 + (38.46)
= 95.82 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00181 318.20 8.84 309.36 174.94 104 18.19
2 565 -0.00581 -360.90 0.00 -360.90 -204.08 -104 21.22
Total-29.15 39.42
xuy = 96 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 96/300 = 0.115
Puc = 0.115 x 20.00 x 450 x 300
= 309.45 kN
Puy1 = Puc + Pus(Total)
= 309.45 + (-29.15)
= 280.30 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 96/300
= 0.132
Muc = 309.45 x (0.5 x 300 - 0.132 x 300) = 34.12 kN-m
Muy1 = Muc + Mus(Total)
= 34.12 + (39.42)
= 73.54 kN-m
Pu/Puz = 0.179
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.179, an = 1.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((58.86/95.82)1.000) + ((15.28/73.54)1.000)
= 0.822 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG6 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 558.27 kN
MomentX,(Mx) = 56.56 kN-m
MomentY,(My) = 15.38 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C140.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 11.556 kN.m
My_MinEccen = 11.165 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 56.563 kN.m
My = max(My,My_MinEccen) + MuaddY = 15.385 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 56.56/558.27 = 101 mm
Actual eccenY = My / P = 15.38/558.27 = 28 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(101,21) = 101 mm
eccenY = max(Actual eccenY,eccenYMin) = max(28,20) = 28 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(101 mm) > 0.05 x 300(15 mm)
and eccenY(28 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00284 352.51 8.92 343.59 77.72 179 13.91
2 226 0.00156 299.98 8.49 291.49 65.93 90 5.90
3 226 0.00028 55.40 2.30 53.10 12.01 0 0.00
4 226 -0.00101 -201.01 0.00 -201.01 -45.47 -90 4.07
5 226 -0.00229 -338.29 0.00 -338.29 -76.52 -179 13.70
Total33.67 37.58
xux = 244 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 244/450 = 0.195
Puc = 0.195 x 20.00 x 300 x 450
= 527.77 kN
Pux1 = Puc + Pus(Total)
= 527.77 + (33.67)
= 561.44 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 244/450
= 0.226
Muc = 527.77 x (0.5 x 450 - 0.226 x 450) = 65.10 kN-m
Mux1 = Muc + Mus(Total)
= 65.10 + (37.58)
= 102.68 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00255 346.30 8.92 337.38 190.78 104 19.84
2 565 -0.00177 -315.32 0.00 -315.32 -178.31 -104 18.54
Total12.47 38.39
xuy = 169 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 169/300 = 0.203
Puc = 0.203 x 20.00 x 450 x 300
= 546.75 kN
Puy1 = Puc + Pus(Total)
= 546.75 + (12.47)
= 559.22 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 169/300
= 0.234
Muc = 546.75 x (0.5 x 300 - 0.234 x 300) = 43.63 kN-m
Muy1 = Muc + Mus(Total)
= 43.63 + (38.39)
= 82.02 kN-m
Pu/Puz = 0.359
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.359, an = 1.264
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((56.56/102.68)1.264) + ((15.38/82.02)1.264)
= 0.591 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG6 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 833.76 kN
MomentX,(Mx) = 48.07 kN-m
MomentY,(My) = 16.91 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C140.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 17.259 kN.m
My_MinEccen = 16.675 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 48.074 kN.m
My = max(My,My_MinEccen) + MuaddY = 16.910 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 48.07/833.76 = 58 mm
Actual eccenY = My / P = 16.91/833.76 = 20 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(58,21) = 58 mm
eccenY = max(Actual eccenY,eccenYMin) = max(20,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(58 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00299 353.82 8.92 344.90 78.02 179 13.96
2 226 0.00200 327.78 8.92 318.86 72.12 90 6.46
3 226 0.00101 202.22 6.74 195.48 44.22 0 0.00
4 226 0.00002 4.22 0.19 4.03 0.91 -90 -0.08
5 226 -0.00097 -193.79 0.00 -193.79 -43.83 -179 7.85
Total151.43 28.18
xux = 316 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 316/450 = 0.253
Puc = 0.253 x 20.00 x 300 x 450
= 683.44 kN
Pux1 = Puc + Pus(Total)
= 683.44 + (151.43)
= 834.87 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 316/450
= 0.293
Muc = 683.44 x (0.5 x 450 - 0.293 x 450) = 63.82 kN-m
Mux1 = Muc + Mus(Total)
= 63.82 + (28.18)
= 92.00 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00276 351.81 8.92 342.89 193.90 104 20.17
2 565 -0.00058 -115.71 0.00 -115.71 -65.43 -104 6.81
Total128.47 26.97
xuy = 218 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 218/300 = 0.262
Puc = 0.262 x 20.00 x 450 x 300
= 706.22 kN
Puy1 = Puc + Pus(Total)
= 706.22 + (128.47)
= 834.69 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 218/300
= 0.302
Muc = 706.22 x (0.5 x 300 - 0.302 x 300) = 41.90 kN-m
Muy1 = Muc + Mus(Total)
= 41.90 + (26.97)
= 68.87 kN-m
Pu/Puz = 0.536
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.536, an = 1.559
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((48.07/92.00)1.559) + ((16.91/68.87)1.559)
= 0.475 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG6 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1148.59 kN
MomentX,(Mx) = 34.53 kN-m
MomentY,(My) = 15.09 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C140.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 23.776 kN.m
My_MinEccen = 22.972 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 34.528 kN.m
My = max(My,My_MinEccen) + MuaddY = 22.972 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 34.53/1148.59 = 30 mm
Actual eccenY = My / P = 15.09/1148.59 = 13 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(30,21) = 30 mm
eccenY = max(Actual eccenY,eccenYMin) = max(13,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(30 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00311 354.88 8.92 345.96 78.25 179 14.01
2 226 0.00236 340.85 8.92 331.93 75.08 90 6.72
3 226 0.00160 304.02 8.57 295.45 66.83 0 0.00
4 226 0.00085 169.32 5.95 163.36 36.95 -90 -3.31
5 226 0.00009 18.30 0.80 17.50 3.96 -179 -0.71
Total261.07 16.71
xux = 415 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 415/450 = 0.332
Puc = 0.332 x 20.00 x 300 x 450
= 896.06 kN
Pux1 = Puc + Pus(Total)
= 896.06 + (261.07)
= 1157.14 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 415/450
= 0.384
Muc = 896.06 x (0.5 x 450 - 0.384 x 450) = 46.98 kN-m
Mux1 = Muc + Mus(Total)
= 46.98 + (16.71)
= 63.69 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00293 353.31 8.92 344.39 194.75 104 20.25
2 565 0.00037 73.05 2.96 70.09 39.63 -104 -4.12
Total234.38 16.13
xuy = 284 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 284/300 = 0.340
Puc = 0.340 x 20.00 x 450 x 300
= 918.84 kN
Puy1 = Puc + Pus(Total)
= 918.84 + (234.38)
= 1153.22 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 284/300
= 0.393
Muc = 918.84 x (0.5 x 300 - 0.393 x 300) = 29.43 kN-m
Muy1 = Muc + Mus(Total)
= 29.43 + (16.13)
= 45.56 kN-m
Pu/Puz = 0.738
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.738, an = 1.896
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((34.53/63.69)1.896) + ((22.97/45.56)1.896)
= 0.586 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG6 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 1379.84 kN
MomentX,(Mx) = 18.18 kN-m
MomentY,(My) = 6.44 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C140.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 30.012 kN.m
My_MinEccen = 27.597 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 30.012 kN.m
My = max(My,My_MinEccen) + MuaddY = 27.597 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 18.18/1379.84 = 13 mm
Actual eccenY = My / P = 6.44/1379.84 = 5 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(13,22) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(5,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 14 nos. (1583 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00313 355.01 8.92 346.09 78.28 179 14.01
2 226 0.00267 349.46 8.92 340.54 77.03 119 9.19
3 226 0.00221 335.50 8.92 326.58 73.87 60 4.41
4 226 0.00175 314.40 8.78 305.62 69.13 -0 -0.00
5 226 0.00130 259.11 7.81 251.30 56.84 -60 -3.39
6 226 0.00084 167.55 5.91 161.64 36.56 -119 -4.36
7 226 0.00038 75.99 3.07 72.92 16.49 -179 -2.95
Total408.21 16.91
xux = 454 mm               Puc = C1.fck.B.D
ku = 1.008
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.112
C1 = 0.446 x (1 - C3/6) = 0.363
Puc = 0.363 x 20.00 x 300 x 450
= 980.97 kN
Pux1 = Puc + Pus(Total)
= 980.97 + (408.21)
= 1389.18 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.419
Muc = 980.97 x (0.5 x 450 - 0.419 x 450) = 35.88 kN-m
Mux1 = Muc + Mus(Total)
= 35.88 + (16.91)
= 52.78 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 792 0.00288 352.84 8.92 343.92 272.28 104 28.32
2 792 0.00066 132.90 4.94 127.95 101.30 -104 -10.54
Total373.58 17.78
xuy = 316 mm               Puc = C1.fck.B.D
ku = 1.055
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.952
C1 = 0.446 x (1 - C3/6) = 0.375
Puc = 0.375 x 20.00 x 450 x 300
= 1013.15 kN
Puy1 = Puc + Pus(Total)
= 1013.15 + (373.58)
= 1386.72 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.433
Muc = 1013.15 x (0.5 x 300 - 0.433 x 300) = 20.47 kN-m
Muy1 = Muc + Mus(Total)
= 20.47 + (17.78)
= 38.25 kN-m
Pu/Puz = 0.815
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.815, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((30.01/52.78)2.000) + ((27.60/38.25)2.000)
= 0.844 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG6 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 14 nos. (1583 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG7 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3450 mm
From analysis results,loads on column
Axial load,(P) = 387.26 kN
MomentX,(Mx) = 127.59 kN-m
MomentY,(My) = 14.59 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C150.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3450 x 1.000 ) / 450 = 7.67 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3450 x 1.000 ) / 300 = 11.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 8.481 kN.m
My_MinEccen = 7.745 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 127.586 kN.m
My = max(My,My_MinEccen) + MuaddY = 14.588 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 127.59/387.26 = 329 mm
Actual eccenY = My / P = 14.59/387.26 = 38 mm
eccenXMin = (L/500) + (D/30) = (3450/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3450/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(329,22) = 329 mm
eccenY = max(Actual eccenY,eccenYMin) = max(38,20) = 38 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(329 mm) > 0.05 x 300(15 mm)
and eccenY(38 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 16 nos. + #16 - 4 nos. (2614 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00270 350.35 8.92 341.43 137.30 177 24.30
2 226 0.00203 328.66 8.92 319.74 72.32 136 9.84
3 226 0.00138 275.97 8.06 267.90 60.60 97 5.89
4 226 0.00073 146.91 5.35 141.56 32.02 58 1.87
5 226 0.00009 17.86 0.78 17.08 3.86 19 0.08
6 226 -0.00056 -111.19 0.00 -111.19 -25.15 -19 0.49
7 226 -0.00120 -240.25 0.00 -240.25 -54.34 -58 3.17
8 226 -0.00185 -320.21 0.00 -320.21 -72.43 -97 7.04
9 226 -0.00249 -344.90 0.00 -344.90 -78.02 -136 10.62
10 402 -0.00317 -355.39 0.00 -355.39 -142.91 -177 25.30
Total-66.75 88.59
xux = 211 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 211/450 = 0.169
Puc = 0.169 x 20.00 x 300 x 450
= 455.63 kN
Pux1 = Puc + Pus(Total)
= 455.63 + (-66.75)
= 388.88 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 211/450
= 0.195
Muc = 455.63 x (0.5 x 450 - 0.195 x 450) = 62.53 kN-m
Mux1 = Muc + Mus(Total)
= 62.53 + (88.59)
= 151.13 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1307 0.00222 335.82 8.92 326.90 427.23 102 43.58
2 1307 -0.00322 -355.83 0.00 -355.83 -465.03 -102 47.43
Total-37.80 91.01
xuy = 131 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 131/300 = 0.157
Puc = 0.157 x 20.00 x 450 x 300
= 425.25 kN
Puy1 = Puc + Pus(Total)
= 425.25 + (-37.80)
= 387.45 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 131/300
= 0.182
Muc = 425.25 x (0.5 x 300 - 0.182 x 300) = 40.57 kN-m
Muy1 = Muc + Mus(Total)
= 40.57 + (91.01)
= 131.58 kN-m
Pu/Puz = 0.193
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.193, an = 1.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((127.59/151.13)1.000) + ((14.59/131.58)1.000)
= 0.955 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG7 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 16 nos. + #16 - 4 nos. (2614 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3450 mm
From analysis results,loads on column
Axial load,(P) = 790.58 kN
MomentX,(Mx) = 112.81 kN-m
MomentY,(My) = 14.18 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C150.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3450 x 1.000 ) / 450 = 7.67 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3450 x 1.000 ) / 300 = 11.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 17.314 kN.m
My_MinEccen = 15.812 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 112.812 kN.m
My = max(My,My_MinEccen) + MuaddY = 15.812 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 112.81/790.58 = 143 mm
Actual eccenY = My / P = 14.18/790.58 = 18 mm
eccenXMin = (L/500) + (D/30) = (3450/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3450/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(143,22) = 143 mm
eccenY = max(Actual eccenY,eccenYMin) = max(18,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(143 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 2 nos. + #16 - 8 nos. (1835 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00292 353.21 8.92 344.29 138.45 177 24.50
2 402 0.00184 319.85 8.86 310.99 125.06 88 10.94
3 226 0.00078 156.97 5.63 151.34 34.23 0 0.00
4 402 -0.00027 -54.21 0.00 -54.21 -21.80 -88 1.91
5 402 -0.00135 -270.21 0.00 -270.21 -108.66 -177 19.23
Total167.28 56.59
xux = 290 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 290/450 = 0.232
Puc = 0.232 x 20.00 x 300 x 450
= 626.48 kN
Pux1 = Puc + Pus(Total)
= 626.48 + (167.28)
= 793.76 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 290/450
= 0.268
Muc = 626.48 x (0.5 x 450 - 0.268 x 450) = 65.37 kN-m
Mux1 = Muc + Mus(Total)
= 65.37 + (56.59)
= 121.96 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 917 0.00266 349.27 8.92 340.35 312.22 102 31.85
2 917 -0.00090 -180.28 0.00 -180.28 -165.38 -102 16.87
Total146.84 48.72
xuy = 200 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 200/300 = 0.240
Puc = 0.240 x 20.00 x 450 x 300
= 649.27 kN
Puy1 = Puc + Pus(Total)
= 649.27 + (146.84)
= 796.10 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 200/300
= 0.278
Muc = 649.27 x (0.5 x 300 - 0.278 x 300) = 43.27 kN-m
Muy1 = Muc + Mus(Total)
= 43.27 + (48.72)
= 91.98 kN-m
Pu/Puz = 0.447
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.447, an = 1.411
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((112.81/121.96)1.411) + ((15.81/91.98)1.411)
= 0.979 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG7 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 2 nos. + #16 - 8 nos. (1835 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3150 mm
From analysis results,loads on column
Axial load,(P) = 1200.61 kN
MomentX,(Mx) = 97.34 kN-m
MomentY,(My) = 9.59 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C150.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3150 x 1.000 ) / 450 = 7.00 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3150 x 1.000 ) / 300 = 10.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 25.573 kN.m
My_MinEccen = 24.012 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 97.343 kN.m
My = max(My,My_MinEccen) + MuaddY = 24.012 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 97.34/1200.61 = 81 mm
Actual eccenY = My / P = 9.59/1200.61 = 8 mm
eccenXMin = (L/500) + (D/30) = (3150/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (3150/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(81,21) = 81 mm
eccenY = max(Actual eccenY,eccenYMin) = max(8,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(81 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 16 nos. + #16 - 4 nos. (2614 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00302 354.05 8.92 345.13 138.79 177 24.57
2 226 0.00261 347.84 8.92 338.92 76.66 136 10.43
3 226 0.00222 335.64 8.92 326.72 73.90 97 7.18
4 226 0.00182 318.81 8.85 309.96 70.11 58 4.09
5 226 0.00143 286.58 8.20 278.38 62.97 19 1.22
6 226 0.00104 208.37 6.87 201.49 45.58 -19 -0.89
7 226 0.00065 130.15 4.86 125.29 28.34 -58 -1.65
8 226 0.00026 51.94 2.17 49.77 11.26 -97 -1.09
9 226 -0.00013 -26.28 0.00 -26.28 -5.94 -136 0.81
10 402 -0.00054 -108.51 0.00 -108.51 -43.64 -177 7.72
Total458.03 52.40
xux = 348 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 348/450 = 0.278
Puc = 0.278 x 20.00 x 300 x 450
= 751.78 kN
Pux1 = Puc + Pus(Total)
= 751.78 + (458.03)
= 1209.81 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 348/450
= 0.322
Muc = 751.78 x (0.5 x 450 - 0.322 x 450) = 60.30 kN-m
Mux1 = Muc + Mus(Total)
= 60.30 + (52.40)
= 112.70 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1307 0.00281 352.24 8.92 343.32 448.69 102 45.77
2 1307 -0.00012 -23.69 0.00 -23.69 -30.96 -102 3.16
Total417.73 48.92
xuy = 244 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 244/300 = 0.293
Puc = 0.293 x 20.00 x 450 x 300
= 789.75 kN
Puy1 = Puc + Pus(Total)
= 789.75 + (417.73)
= 1207.48 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 244/300
= 0.338
Muc = 789.75 x (0.5 x 300 - 0.338 x 300) = 38.38 kN-m
Muy1 = Muc + Mus(Total)
= 38.38 + (48.92)
= 87.31 kN-m
Pu/Puz = 0.599
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.599, an = 1.665
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((97.34/112.70)1.665) + ((24.01/87.31)1.665)
= 0.900 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG7 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 16 nos. + #16 - 4 nos. (2614 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3150 mm
From analysis results,loads on column
Axial load,(P) = 1779.89 kN
MomentX,(Mx) = 92.72 kN-m
MomentY,(My) = 5.20 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C150.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3150 x 1.000 ) / 450 = 7.00 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3150 x 1.000 ) / 300 = 10.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 37.912 kN.m
My_MinEccen = 35.598 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 92.722 kN.m
My = max(My,My_MinEccen) + MuaddY = 35.598 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 92.72/1779.89 = 52 mm
Actual eccenY = My / P = 5.20/1779.89 = 3 mm
eccenXMin = (L/500) + (D/30) = (3150/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (3150/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(52,21) = 52 mm
eccenY = max(Actual eccenY,eccenYMin) = max(3,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(52 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 14 nos. (4398 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00305 354.32 8.92 345.40 217.02 175 37.98
2 628 0.00252 345.62 8.92 336.70 211.55 117 24.68
3 628 0.00199 327.43 8.92 318.51 200.13 58 11.67
4 628 0.00146 290.94 8.28 282.66 177.60 0 0.00
5 628 0.00094 187.14 6.39 180.74 113.56 -58 -6.62
6 628 0.00041 81.55 3.27 78.28 49.19 -117 -5.74
7 628 -0.00012 -24.04 0.00 -24.04 -15.11 -175 2.64
Total953.94 64.61
xux = 387 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 387/450 = 0.309
Puc = 0.309 x 20.00 x 300 x 450
= 835.31 kN
Pux1 = Puc + Pus(Total)
= 835.31 + (953.94)
= 1789.26 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 387/450
= 0.358
Muc = 835.31 x (0.5 x 450 - 0.358 x 450) = 53.56 kN-m
Mux1 = Muc + Mus(Total)
= 53.56 + (64.61)
= 118.18 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 2199 0.00287 352.74 8.92 343.82 756.10 100 75.61
2 2199 0.00034 67.23 2.75 64.49 141.81 -100 -14.18
Total897.91 61.43
xuy = 277 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 277/300 = 0.332
Puc = 0.332 x 20.00 x 450 x 300
= 896.06 kN
Puy1 = Puc + Pus(Total)
= 896.06 + (897.91)
= 1793.97 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 277/300
= 0.384
Muc = 896.06 x (0.5 x 300 - 0.384 x 300) = 31.32 kN-m
Muy1 = Muc + Mus(Total)
= 31.32 + (61.43)
= 92.75 kN-m
Pu/Puz = 0.700
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.700, an = 1.833
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((92.72/118.18)1.833) + ((35.60/92.75)1.833)
= 0.814 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 20 = 320 mm
               iii) 300 mm
               Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG7 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 14 nos. (4398 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3675 mm
From analysis results,loads on column
Axial load,(P) = 2208.46 kN
MomentX,(Mx) = 76.81 kN-m
MomentY,(My) = 5.73 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C150.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3675 x 1.000 ) / 450 = 8.17 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3675 x 1.000 ) / 300 = 12.25 > 12.00,
column is slender in this direction.
Puz = (0.45 x fck x Ag) + (0.75 x fy - 0.45 x fck) x Ast
= 0.45 x 20.00 x 135000.01 + (0.75 x 415.00 - 0.45 x 20.00 ) x 5890.49 = 2995399.658 N
k1 = 0.190000
k2 = 0.012000 MPa
Pub = (k1 x fck x Ag) + (k2 x Ast x 100)
= (0.190000 x 20.00 x 135000.01) + (0.01 x 5890.49 x 100) = 520068.604 N
k = (Puz - Pu) / (Puz - Pub) <= 1
= (2995399.658 - 2208460.449) / (2995399.658 - 520068.604) = 0.317913
MuaddY = Pu x (Width/2000) x (Lo/Width) x (Lo/Width) x k
= 2208460.45 x (300.00/2000) x (3675.00/300.00) x (3675.00/300.00) x 0.317913
= 15803779.602 N.mm = 15.804 kN.m
Mx_MinEccen = 49.359 kN.m
My_MinEccen = 44.169 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 76.806 kN.m
My = max(My,My_MinEccen) + MuaddY = 59.973 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 76.81/2208.46 = 35 mm
Actual eccenY = My / P = 5.73/2208.46 = 3 mm
eccenXMin = (L/500) + (D/30) = (3675/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3675/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(35,22) = 35 mm
eccenY = max(Actual eccenY,eccenYMin) = max(3,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(35 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #25 - 12 nos. (5890 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 982 0.00305 354.30 8.92 345.38 339.07 173 58.49
2 982 0.00245 343.78 8.92 334.86 328.75 104 34.03
3 982 0.00185 320.48 8.87 311.61 305.93 35 10.55
4 982 0.00125 250.70 7.68 243.02 238.59 -35 -8.23
5 982 0.00066 131.23 4.89 126.34 124.04 -104 -12.84
6 982 0.00006 11.77 0.52 11.25 11.05 -173 -1.91
Total1347.42 80.10
xux = 404 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 404/450 = 0.323
Puc = 0.323 x 20.00 x 300 x 450
= 873.28 kN
Pux1 = Puc + Pus(Total)
= 873.28 + (1347.42)
= 2220.70 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 404/450
= 0.374
Muc = 873.28 x (0.5 x 450 - 0.374 x 450) = 49.61 kN-m
Mux1 = Muc + Mus(Total)
= 49.61 + (80.10)
= 129.71 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 2945 0.00286 352.70 8.92 343.78 1012.51 98 98.72
2 2945 0.00050 99.02 3.87 95.15 280.25 -98 -27.32
Total1292.76 71.39
xuy = 288 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 288/300 = 0.346
Puc = 0.346 x 20.00 x 450 x 300
= 934.03 kN
Puy1 = Puc + Pus(Total)
= 934.03 + (1292.76)
= 2226.79 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 288/300
= 0.400
Muc = 934.03 x (0.5 x 300 - 0.400 x 300) = 28.09 kN-m
Muy1 = Muc + Mus(Total)
= 28.09 + (71.39)
= 99.49 kN-m
Pu/Puz = 0.737
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.737, an = 1.895
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((76.81/129.71)1.895) + ((59.97/99.49)1.895)
= 0.754 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 25/4 = 6 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (6, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 25 = 400 mm
               iii) 300 mm
               Required spacing = minimum of (300, 400, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG7 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #25 - 12 nos. (5890 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG8 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3450 mm
From analysis results,loads on column
Axial load,(P) = 261.01 kN
MomentX,(Mx) = 152.50 kN-m
MomentY,(My) = 12.19 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C160.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3450 x 1.000 ) / 450 = 7.67 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3450 x 1.000 ) / 300 = 11.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 5.716 kN.m
My_MinEccen = 5.220 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 152.503 kN.m
My = max(My,My_MinEccen) + MuaddY = 12.187 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 152.50/261.01 = 584 mm
Actual eccenY = My / P = 12.19/261.01 = 47 mm
eccenXMin = (L/500) + (D/30) = (3450/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3450/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(584,22) = 584 mm
eccenY = max(Actual eccenY,eccenYMin) = max(47,20) = 47 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(584 mm) > 0.05 x 300(15 mm)
and eccenY(47 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 18 nos. (3619 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00266 349.27 8.92 340.35 136.86 177 24.22
2 402 0.00189 322.85 8.89 313.96 126.25 133 16.76
3 402 0.00112 223.18 7.18 216.00 86.86 89 7.69
4 402 0.00034 68.61 2.80 65.81 26.46 44 1.17
5 402 -0.00043 -85.96 0.00 -85.96 -34.57 0 0.00
6 402 -0.00120 -240.54 0.00 -240.54 -96.73 -44 4.28
7 402 -0.00198 -326.84 0.00 -326.84 -131.43 -89 11.63
8 402 -0.00275 -351.50 0.00 -351.50 -141.35 -133 18.76
9 402 -0.00352 -358.46 0.00 -358.46 -144.15 -177 25.51
Total-171.78 110.03
xux = 200 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 200/450 = 0.160
Puc = 0.160 x 20.00 x 300 x 450
= 432.84 kN
Pux1 = Puc + Pus(Total)
= 432.84 + (-171.78)
= 261.06 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 200/450
= 0.185
Muc = 432.84 x (0.5 x 450 - 0.185 x 450) = 61.31 kN-m
Mux1 = Muc + Mus(Total)
= 61.31 + (110.03)
= 171.34 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1810 0.00192 324.54 8.90 315.64 571.17 102 58.26
2 1810 -0.00482 -360.90 0.00 -360.90 -653.07 -102 66.61
Total-81.90 124.87
xuy = 106 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 106/300 = 0.127
Puc = 0.127 x 20.00 x 450 x 300
= 343.62 kN
Puy1 = Puc + Pus(Total)
= 343.62 + (-81.90)
= 261.72 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 106/300
= 0.147
Muc = 343.62 x (0.5 x 300 - 0.147 x 300) = 36.38 kN-m
Muy1 = Muc + Mus(Total)
= 36.38 + (124.87)
= 161.26 kN-m
Pu/Puz = 0.113
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.113, an = 1.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((152.50/171.34)1.000) + ((12.19/161.26)1.000)
= 0.966 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG8 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 18 nos. (3619 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3450 mm
From analysis results,loads on column
Axial load,(P) = 541.65 kN
MomentX,(Mx) = 98.60 kN-m
MomentY,(My) = 9.32 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C160.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3450 x 1.000 ) / 450 = 7.67 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3450 x 1.000 ) / 300 = 11.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 11.862 kN.m
My_MinEccen = 10.833 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 98.603 kN.m
My = max(My,My_MinEccen) + MuaddY = 10.833 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 98.60/541.65 = 182 mm
Actual eccenY = My / P = 9.32/541.65 = 17 mm
eccenXMin = (L/500) + (D/30) = (3450/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3450/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(182,22) = 182 mm
eccenY = max(Actual eccenY,eccenYMin) = max(17,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(182 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 6 nos. (1206 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00281 352.22 8.92 343.30 138.05 177 24.43
2 402 0.00025 50.72 2.12 48.61 19.55 0 0.00
3 402 -0.00230 -338.77 0.00 -338.77 -136.23 -177 24.11
Total21.37 48.55
xux = 243 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 243/450 = 0.194
Puc = 0.194 x 20.00 x 300 x 450
= 523.97 kN
Pux1 = Puc + Pus(Total)
= 523.97 + (21.37)
= 545.34 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 243/450
= 0.224
Muc = 523.97 x (0.5 x 450 - 0.224 x 450) = 65.02 kN-m
Mux1 = Muc + Mus(Total)
= 65.02 + (48.55)
= 113.56 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 603 0.00248 344.68 8.92 335.76 202.53 102 20.66
2 603 -0.00184 -319.67 0.00 -319.67 -192.82 -102 19.67
Total9.71 40.33
xuy = 165 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 165/300 = 0.198
Puc = 0.198 x 20.00 x 450 x 300
= 535.36 kN
Puy1 = Puc + Pus(Total)
= 535.36 + (9.71)
= 545.06 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 165/300
= 0.229
Muc = 535.36 x (0.5 x 300 - 0.229 x 300) = 43.50 kN-m
Muy1 = Muc + Mus(Total)
= 43.50 + (40.33)
= 83.83 kN-m
Pu/Puz = 0.343
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.343, an = 1.238
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((98.60/113.56)1.238) + ((10.83/83.83)1.238)
= 0.919 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG8 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 6 nos. (1206 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3150 mm
From analysis results,loads on column
Axial load,(P) = 815.23 kN
MomentX,(Mx) = 90.57 kN-m
MomentY,(My) = 6.11 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C160.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3150 x 1.000 ) / 450 = 7.00 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3150 x 1.000 ) / 300 = 10.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 17.365 kN.m
My_MinEccen = 16.305 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 90.567 kN.m
My = max(My,My_MinEccen) + MuaddY = 16.305 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 90.57/815.23 = 111 mm
Actual eccenY = My / P = 6.11/815.23 = 7 mm
eccenXMin = (L/500) + (D/30) = (3150/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (3150/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(111,21) = 111 mm
eccenY = max(Actual eccenY,eccenYMin) = max(7,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(111 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 6 nos. (1206 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00297 353.60 8.92 344.68 138.61 177 24.53
2 402 0.00100 199.44 6.68 192.76 77.51 0 0.00
3 402 -0.00097 -194.33 0.00 -194.33 -78.15 -177 13.83
Total137.97 38.36
xux = 315 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 315/450 = 0.252
Puc = 0.252 x 20.00 x 300 x 450
= 679.64 kN
Pux1 = Puc + Pus(Total)
= 679.64 + (137.97)
= 817.62 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 315/450
= 0.291
Muc = 679.64 x (0.5 x 450 - 0.291 x 450) = 63.96 kN-m
Mux1 = Muc + Mus(Total)
= 63.96 + (38.36)
= 102.32 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 603 0.00271 350.57 8.92 341.65 206.08 102 21.02
2 603 -0.00064 -127.08 0.00 -127.08 -76.65 -102 7.82
Total129.43 28.84
xuy = 213 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 213/300 = 0.256
Puc = 0.256 x 20.00 x 450 x 300
= 691.03 kN
Puy1 = Puc + Pus(Total)
= 691.03 + (129.43)
= 820.46 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 213/300
= 0.296
Muc = 691.03 x (0.5 x 300 - 0.296 x 300) = 42.34 kN-m
Muy1 = Muc + Mus(Total)
= 42.34 + (28.84)
= 71.18 kN-m
Pu/Puz = 0.516
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.516, an = 1.527
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((90.57/102.32)1.527) + ((16.30/71.18)1.527)
= 0.935 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG8 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 6 nos. (1206 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3150 mm
From analysis results,loads on column
Axial load,(P) = 1562.72 kN
MomentX,(Mx) = 91.77 kN-m
MomentY,(My) = 5.99 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C160.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3150 x 1.000 ) / 450 = 7.00 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3150 x 1.000 ) / 300 = 10.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 33.286 kN.m
My_MinEccen = 31.254 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 91.773 kN.m
My = max(My,My_MinEccen) + MuaddY = 31.254 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 91.77/1562.72 = 59 mm
Actual eccenY = My / P = 5.99/1562.72 = 4 mm
eccenXMin = (L/500) + (D/30) = (3150/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (3150/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(59,21) = 59 mm
eccenY = max(Actual eccenY,eccenYMin) = max(4,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(59 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 18 nos. (3619 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00305 354.37 8.92 345.45 138.91 177 24.59
2 402 0.00264 348.76 8.92 339.84 136.66 133 18.14
3 402 0.00223 336.19 8.92 327.27 131.60 89 11.65
4 402 0.00182 318.45 8.85 309.60 124.50 44 5.51
5 402 0.00141 281.31 8.13 273.17 109.85 0 0.00
6 402 0.00099 198.97 6.67 192.30 77.33 -44 -3.42
7 402 0.00058 116.62 4.44 112.18 45.11 -89 -3.99
8 402 0.00017 34.28 1.46 32.82 13.20 -133 -1.75
9 402 -0.00024 -48.06 0.00 -48.06 -19.33 -177 3.42
Total757.83 54.14
xux = 376 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 376/450 = 0.301
Puc = 0.301 x 20.00 x 300 x 450
= 812.53 kN
Pux1 = Puc + Pus(Total)
= 812.53 + (757.83)
= 1570.36 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 376/450
= 0.348
Muc = 812.53 x (0.5 x 450 - 0.348 x 450) = 55.67 kN-m
Mux1 = Muc + Mus(Total)
= 55.67 + (54.14)
= 109.81 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1810 0.00288 352.82 8.92 343.90 622.31 102 63.48
2 1810 0.00023 45.53 1.92 43.62 78.92 -102 -8.05
Total701.23 55.43
xuy = 270 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 270/300 = 0.323
Puc = 0.323 x 20.00 x 450 x 300
= 873.28 kN
Puy1 = Puc + Pus(Total)
= 873.28 + (701.23)
= 1574.51 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 270/300
= 0.374
Muc = 873.28 x (0.5 x 300 - 0.374 x 300) = 33.08 kN-m
Muy1 = Muc + Mus(Total)
= 33.08 + (55.43)
= 88.50 kN-m
Pu/Puz = 0.677
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.677, an = 1.795
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((91.77/109.81)1.795) + ((31.25/88.50)1.795)
= 0.879 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG8 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 18 nos. (3619 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3675 mm
From analysis results,loads on column
Axial load,(P) = 1819.48 kN
MomentX,(Mx) = 77.39 kN-m
MomentY,(My) = 5.52 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C160.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3675 x 1.000 ) / 450 = 8.17 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3675 x 1.000 ) / 300 = 12.25 > 12.00,
column is slender in this direction.
Puz = (0.45 x fck x Ag) + (0.75 x fy - 0.45 x fck) x Ast
= 0.45 x 20.00 x 135000.01 + (0.75 x 415.00 - 0.45 x 20.00 ) x 4398.23 = 2544364.990 N
k1 = 0.192000
k2 = 0.023333 MPa
Pub = (k1 x fck x Ag) + (k2 x Ast x 100)
= (0.192000 x 20.00 x 135000.01) + (0.02 x 4398.23 x 100) = 528662.598 N
k = (Puz - Pu) / (Puz - Pub) <= 1
= (2544364.990 - 1819484.985) / (2544364.990 - 528662.598) = 0.359617
MuaddY = Pu x (Width/2000) x (Lo/Width) x (Lo/Width) x k
= 1819484.99 x (300.00/2000) x (3675.00/300.00) x (3675.00/300.00) x 0.359617
= 14728266.716 N.mm = 14.728 kN.m
Mx_MinEccen = 40.665 kN.m
My_MinEccen = 36.390 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 77.386 kN.m
My = max(My,My_MinEccen) + MuaddY = 51.118 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 77.39/1819.48 = 43 mm
Actual eccenY = My / P = 5.52/1819.48 = 3 mm
eccenXMin = (L/500) + (D/30) = (3675/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3675/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(43,22) = 43 mm
eccenY = max(Actual eccenY,eccenYMin) = max(3,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(43 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 14 nos. (4398 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00306 354.39 8.92 345.47 217.06 175 37.99
2 628 0.00254 346.07 8.92 337.15 211.84 117 24.71
3 628 0.00202 328.42 8.92 319.50 200.75 58 11.71
4 628 0.00150 294.38 8.36 286.02 179.71 0 0.00
5 628 0.00098 196.30 6.61 189.69 119.19 -58 -6.95
6 628 0.00046 92.59 3.65 88.94 55.88 -117 -6.52
7 628 -0.00006 -11.11 0.00 -11.11 -6.98 -175 1.22
Total977.45 62.16
xux = 394 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 394/450 = 0.315
Puc = 0.315 x 20.00 x 300 x 450
= 850.50 kN
Pux1 = Puc + Pus(Total)
= 850.50 + (977.45)
= 1827.95 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 394/450
= 0.364
Muc = 850.50 x (0.5 x 450 - 0.364 x 450) = 52.05 kN-m
Mux1 = Muc + Mus(Total)
= 52.05 + (62.16)
= 114.21 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 2199 0.00288 352.83 8.92 343.91 756.30 100 75.63
2 2199 0.00039 77.78 3.13 74.65 164.16 -100 -16.42
Total920.45 59.21
xuy = 281 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 281/300 = 0.338
Puc = 0.338 x 20.00 x 450 x 300
= 911.25 kN
Puy1 = Puc + Pus(Total)
= 911.25 + (920.45)
= 1831.70 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 281/300
= 0.390
Muc = 911.25 x (0.5 x 300 - 0.390 x 300) = 30.07 kN-m
Muy1 = Muc + Mus(Total)
= 30.07 + (59.21)
= 89.29 kN-m
Pu/Puz = 0.715
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.715, an = 1.859
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((77.39/114.21)1.859) + ((51.12/89.29)1.859)
= 0.840 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 20 = 320 mm
               iii) 300 mm
               Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG8 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 14 nos. (4398 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG9 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 305.84 kN
MomentX,(Mx) = 3.88 kN-m
MomentY,(My) = 59.81 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C170.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 6.331 kN.m
My_MinEccen = 6.117 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 6.331 kN.m
My = max(My,My_MinEccen) + MuaddY = 59.814 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 3.88/305.84 = 13 mm
Actual eccenY = My / P = 59.81/305.84 = 196 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(13,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(196,20) = 196 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(196 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00261 347.96 8.92 339.04 76.69 179 13.73
2 226 0.00088 176.13 6.13 170.00 38.45 90 3.44
3 226 -0.00085 -169.90 0.00 -169.90 -38.43 0 0.00
4 226 -0.00258 -347.16 0.00 -347.16 -78.53 -90 7.03
5 226 -0.00431 -360.90 0.00 -360.90 -81.63 -179 14.61
Total-83.45 38.81
xux = 181 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 181/450 = 0.145
Puc = 0.145 x 20.00 x 300 x 450
= 391.08 kN
Pux1 = Puc + Pus(Total)
= 391.08 + (-83.45)
= 307.63 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 181/450
= 0.167
Muc = 391.08 x (0.5 x 450 - 0.167 x 450) = 58.54 kN-m
Mux1 = Muc + Mus(Total)
= 58.54 + (38.81)
= 97.35 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00193 325.16 8.91 316.25 178.84 104 18.60
2 565 -0.00517 -360.90 0.00 -360.90 -204.08 -104 21.22
Total-25.25 39.82
xuy = 103 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 103/300 = 0.123
Puc = 0.123 x 20.00 x 450 x 300
= 332.23 kN
Puy1 = Puc + Pus(Total)
= 332.23 + (-25.25)
= 306.98 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 103/300
= 0.142
Muc = 332.23 x (0.5 x 300 - 0.142 x 300) = 35.66 kN-m
Muy1 = Muc + Mus(Total)
= 35.66 + (39.82)
= 75.49 kN-m
Pu/Puz = 0.196
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.196, an = 1.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((6.33/97.35)1.000) + ((59.81/75.49)1.000)
= 0.857 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG9 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 615.31 kN
MomentX,(Mx) = 3.32 kN-m
MomentY,(My) = 54.52 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C170.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 12.737 kN.m
My_MinEccen = 12.306 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 12.737 kN.m
My = max(My,My_MinEccen) + MuaddY = 54.518 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 3.32/615.31 = 5 mm
Actual eccenY = My / P = 54.52/615.31 = 89 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(5,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(89,20) = 89 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(89 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00288 352.84 8.92 343.92 77.79 179 13.92
2 226 0.00167 309.25 8.68 300.57 67.99 90 6.08
3 226 0.00046 92.54 3.65 88.89 20.11 0 0.00
4 226 -0.00075 -149.09 0.00 -149.09 -33.72 -90 3.02
5 226 -0.00195 -326.04 0.00 -326.04 -73.75 -179 13.20
Total58.42 36.23
xux = 259 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 259/450 = 0.207
Puc = 0.207 x 20.00 x 300 x 450
= 560.04 kN
Pux1 = Puc + Pus(Total)
= 560.04 + (58.42)
= 618.46 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 259/450
= 0.240
Muc = 560.04 x (0.5 x 450 - 0.240 x 450) = 65.60 kN-m
Mux1 = Muc + Mus(Total)
= 65.60 + (36.23)
= 101.83 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00261 347.96 8.92 339.04 191.72 104 19.94
2 565 -0.00141 -282.02 0.00 -282.02 -159.48 -104 16.59
Total32.24 36.53
xuy = 181 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 181/300 = 0.217
Puc = 0.217 x 20.00 x 450 x 300
= 586.62 kN
Puy1 = Puc + Pus(Total)
= 586.62 + (32.24)
= 618.86 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 181/300
= 0.251
Muc = 586.62 x (0.5 x 300 - 0.251 x 300) = 43.81 kN-m
Muy1 = Muc + Mus(Total)
= 43.81 + (36.53)
= 80.33 kN-m
Pu/Puz = 0.395
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.395, an = 1.325
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((12.74/101.83)1.325) + ((54.52/80.33)1.325)
= 0.662 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG9 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 933.62 kN
MomentX,(Mx) = 3.66 kN-m
MomentY,(My) = 53.03 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C170.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 19.326 kN.m
My_MinEccen = 18.672 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 19.326 kN.m
My = max(My,My_MinEccen) + MuaddY = 53.028 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 3.66/933.62 = 4 mm
Actual eccenY = My / P = 53.03/933.62 = 57 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(4,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(57,20) = 57 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(57 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00303 354.19 8.92 345.27 78.10 179 13.98
2 226 0.00212 332.28 8.92 323.36 73.14 90 6.55
3 226 0.00121 242.86 7.54 235.31 53.23 0 0.00
4 226 0.00031 61.02 2.51 58.50 13.23 -90 -1.18
5 226 -0.00060 -120.83 0.00 -120.83 -27.33 -179 4.89
Total190.37 24.23
xux = 345 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 345/450 = 0.276
Puc = 0.276 x 20.00 x 300 x 450
= 744.19 kN
Pux1 = Puc + Pus(Total)
= 744.19 + (190.37)
= 934.56 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 345/450
= 0.319
Muc = 744.19 x (0.5 x 450 - 0.319 x 450) = 60.78 kN-m
Mux1 = Muc + Mus(Total)
= 60.78 + (24.23)
= 85.01 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00282 352.35 8.92 343.43 194.21 104 20.20
2 565 -0.00024 -47.40 0.00 -47.40 -26.81 -104 2.79
Total167.40 22.99
xuy = 238 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 238/300 = 0.285
Puc = 0.285 x 20.00 x 450 x 300
= 770.77 kN
Puy1 = Puc + Pus(Total)
= 770.77 + (167.40)
= 938.17 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 238/300
= 0.330
Muc = 770.77 x (0.5 x 300 - 0.330 x 300) = 39.34 kN-m
Muy1 = Muc + Mus(Total)
= 39.34 + (22.99)
= 62.32 kN-m
Pu/Puz = 0.600
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.600, an = 1.666
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((19.33/85.01)1.666) + ((53.03/62.32)1.666)
= 0.849 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG9 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1265.52 kN
MomentX,(Mx) = 3.97 kN-m
MomentY,(My) = 51.12 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C170.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 26.196 kN.m
My_MinEccen = 25.310 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 26.196 kN.m
My = max(My,My_MinEccen) + MuaddY = 51.117 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 3.97/1265.52 = 3 mm
Actual eccenY = My / P = 51.12/1265.52 = 40 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(3,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(40,20) = 40 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(40 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 16 nos. (1810 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00310 354.76 8.92 345.84 78.23 179 14.00
2 226 0.00265 349.01 8.92 340.09 76.93 128 9.84
3 226 0.00221 335.27 8.92 326.35 73.82 77 5.66
4 226 0.00176 314.71 8.79 305.92 69.20 26 1.77
5 226 0.00131 262.35 7.86 254.49 57.56 -26 -1.47
6 226 0.00087 173.03 6.05 166.98 37.77 -77 -2.90
7 226 0.00042 83.70 3.34 80.36 18.18 -128 -2.32
8 226 -0.00003 -5.62 0.00 -5.62 -1.27 -179 0.23
Total410.41 24.80
xux = 401 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 401/450 = 0.321
Puc = 0.321 x 20.00 x 300 x 450
= 865.69 kN
Pux1 = Puc + Pus(Total)
= 865.69 + (410.41)
= 1276.10 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 401/450
= 0.370
Muc = 865.69 x (0.5 x 450 - 0.370 x 450) = 50.45 kN-m
Mux1 = Muc + Mus(Total)
= 50.45 + (24.80)
= 75.25 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 905 0.00292 353.22 8.92 344.30 311.52 104 32.40
2 905 0.00031 62.51 2.57 59.94 54.23 -104 -5.64
Total365.75 26.76
xuy = 279 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 279/300 = 0.335
Puc = 0.335 x 20.00 x 450 x 300
= 903.66 kN
Puy1 = Puc + Pus(Total)
= 903.66 + (365.75)
= 1269.41 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 279/300
= 0.387
Muc = 903.66 x (0.5 x 300 - 0.387 x 300) = 30.70 kN-m
Muy1 = Muc + Mus(Total)
= 30.70 + (26.76)
= 57.46 kN-m
Pu/Puz = 0.718
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.718, an = 1.864
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((26.20/75.25)1.864) + ((51.12/57.46)1.864)
= 0.944 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG9 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 16 nos. (1810 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 1620.97 kN
MomentX,(Mx) = 2.06 kN-m
MomentY,(My) = 34.62 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C170.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 35.256 kN.m
My_MinEccen = 32.419 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 35.256 kN.m
My = max(My,My_MinEccen) + MuaddY = 34.619 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 2.06/1620.97 = 1 mm
Actual eccenY = My / P = 34.62/1620.97 = 21 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(1,22) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(21,20) = 21 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(21 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 14 nos. (2815 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00311 354.88 8.92 345.96 139.12 177 24.62
2 402 0.00263 348.56 8.92 339.64 136.58 118 16.12
3 402 0.00216 333.48 8.92 324.56 130.52 59 7.70
4 402 0.00168 309.75 8.69 301.06 121.06 0 0.00
5 402 0.00120 240.26 7.50 232.77 93.60 -59 -5.52
6 402 0.00072 144.76 5.29 139.47 56.08 -118 -6.62
7 402 0.00025 49.25 2.06 47.19 18.97 -177 -3.36
Total695.93 32.94
xux = 432 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 432/450 = 0.346
Puc = 0.346 x 20.00 x 300 x 450
= 934.03 kN
Pux1 = Puc + Pus(Total)
= 934.03 + (695.93)
= 1629.96 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 432/450
= 0.400
Muc = 934.03 x (0.5 x 450 - 0.400 x 450) = 42.14 kN-m
Mux1 = Muc + Mus(Total)
= 42.14 + (32.94)
= 75.08 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1407 0.00291 353.16 8.92 344.24 484.49 102 49.42
2 1407 0.00060 119.67 4.54 115.13 162.03 -102 -16.53
Total646.52 32.89
xuy = 305 mm               Puc = C1.fck.B.D
ku = 1.016
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.083
C1 = 0.446 x (1 - C3/6) = 0.366
Puc = 0.366 x 20.00 x 450 x 300
= 986.88 kN
Puy1 = Puc + Pus(Total)
= 986.88 + (646.52)
= 1633.40 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.421
Muc = 986.88 x (0.5 x 300 - 0.421 x 300) = 23.28 kN-m
Muy1 = Muc + Mus(Total)
= 23.28 + (32.89)
= 56.18 kN-m
Pu/Puz = 0.785
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.785, an = 1.974
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((35.26/75.08)1.974) + ((34.62/56.18)1.974)
= 0.609 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG9 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 14 nos. (2815 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG10 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 428.89 kN
MomentX,(Mx) = 58.03 kN-m
MomentY,(My) = 37.49 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C180.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 8.878 kN.m
My_MinEccen = 8.578 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 58.031 kN.m
My = max(My,My_MinEccen) + MuaddY = 37.495 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 58.03/428.89 = 135 mm
Actual eccenY = My / P = 37.49/428.89 = 87 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(135,21) = 135 mm
eccenY = max(Actual eccenY,eccenYMin) = max(87,20) = 87 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(135 mm) > 0.05 x 300(15 mm)
and eccenY(87 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00274 351.36 8.92 342.44 77.46 179 13.87
2 226 0.00127 254.06 7.73 246.32 55.72 90 4.99
3 226 -0.00020 -40.50 0.00 -40.50 -9.16 0 0.00
4 226 -0.00168 -309.52 0.00 -309.52 -70.01 -90 6.27
5 226 -0.00315 -355.20 0.00 -355.20 -80.34 -179 14.38
Total-26.34 39.50
xux = 213 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 213/450 = 0.170
Puc = 0.170 x 20.00 x 300 x 450
= 459.42 kN
Pux1 = Puc + Pus(Total)
= 459.42 + (-26.34)
= 433.08 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 213/450
= 0.197
Muc = 459.42 x (0.5 x 450 - 0.197 x 450) = 62.72 kN-m
Mux1 = Muc + Mus(Total)
= 62.72 + (39.50)
= 102.22 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00233 339.71 8.92 330.79 187.05 104 19.45
2 565 -0.00298 -353.76 0.00 -353.76 -200.05 -104 20.80
Total-12.99 40.26
xuy = 137 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 137/300 = 0.165
Puc = 0.165 x 20.00 x 450 x 300
= 444.23 kN
Puy1 = Puc + Pus(Total)
= 444.23 + (-12.99)
= 431.24 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 137/300
= 0.190
Muc = 444.23 x (0.5 x 300 - 0.190 x 300) = 41.30 kN-m
Muy1 = Muc + Mus(Total)
= 41.30 + (40.26)
= 81.56 kN-m
Pu/Puz = 0.275
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.275, an = 1.126
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((58.03/102.22)1.126) + ((37.49/81.56)1.126)
= 0.946 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG10 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 843.23 kN
MomentX,(Mx) = 55.96 kN-m
MomentY,(My) = 33.28 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C180.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 17.455 kN.m
My_MinEccen = 16.865 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 55.962 kN.m
My = max(My,My_MinEccen) + MuaddY = 33.275 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 55.96/843.23 = 66 mm
Actual eccenY = My / P = 33.28/843.23 = 39 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(66,21) = 66 mm
eccenY = max(Actual eccenY,eccenYMin) = max(39,20) = 39 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(66 mm) > 0.05 x 300(15 mm)
and eccenY(39 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00300 353.87 8.92 344.95 78.03 179 13.97
2 226 0.00202 328.39 8.92 319.47 72.26 90 6.47
3 226 0.00104 207.69 6.86 200.83 45.43 0 0.00
4 226 0.00006 11.86 0.52 11.34 2.57 -90 -0.23
5 226 -0.00092 -183.97 0.00 -183.97 -41.61 -179 7.45
Total156.67 27.65
xux = 320 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 320/450 = 0.256
Puc = 0.256 x 20.00 x 300 x 450
= 691.03 kN
Pux1 = Puc + Pus(Total)
= 691.03 + (156.67)
= 847.70 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 320/450
= 0.296
Muc = 691.03 x (0.5 x 450 - 0.296 x 450) = 63.51 kN-m
Mux1 = Muc + Mus(Total)
= 63.51 + (27.65)
= 91.17 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00277 351.88 8.92 342.96 193.94 104 20.17
2 565 -0.00054 -107.04 0.00 -107.04 -60.53 -104 6.29
Total133.41 26.46
xuy = 220 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 220/300 = 0.264
Puc = 0.264 x 20.00 x 450 x 300
= 713.81 kN
Puy1 = Puc + Pus(Total)
= 713.81 + (133.41)
= 847.23 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 220/300
= 0.306
Muc = 713.81 x (0.5 x 300 - 0.306 x 300) = 41.65 kN-m
Muy1 = Muc + Mus(Total)
= 41.65 + (26.46)
= 68.12 kN-m
Pu/Puz = 0.542
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.542, an = 1.569
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((55.96/91.17)1.569) + ((33.28/68.12)1.569)
= 0.790 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG10 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1264.49 kN
MomentX,(Mx) = 49.63 kN-m
MomentY,(My) = 33.41 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C180.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 26.175 kN.m
My_MinEccen = 25.290 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 49.629 kN.m
My = max(My,My_MinEccen) + MuaddY = 33.410 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 49.63/1264.49 = 39 mm
Actual eccenY = My / P = 33.41/1264.49 = 26 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(39,21) = 39 mm
eccenY = max(Actual eccenY,eccenYMin) = max(26,20) = 26 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(39 mm) > 0.05 x 300(15 mm)
and eccenY(26 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 8 nos. (1608 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00310 354.76 8.92 345.84 139.07 177 24.62
2 402 0.00211 331.83 8.92 322.91 129.85 59 7.66
3 402 0.00112 224.81 7.21 217.60 87.50 -59 -5.16
4 402 0.00014 27.37 1.18 26.19 10.53 -177 -1.86
Total366.95 25.25
xux = 418 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 418/450 = 0.335
Puc = 0.335 x 20.00 x 300 x 450
= 903.66 kN
Pux1 = Puc + Pus(Total)
= 903.66 + (366.95)
= 1270.61 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 418/450
= 0.387
Muc = 903.66 x (0.5 x 450 - 0.387 x 450) = 46.05 kN-m
Mux1 = Muc + Mus(Total)
= 46.05 + (25.25)
= 71.30 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 804 0.00291 353.13 8.92 344.21 276.83 102 28.24
2 804 0.00042 83.08 3.32 79.76 64.15 -102 -6.54
Total340.98 21.69
xuy = 286 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 286/300 = 0.343
Puc = 0.343 x 20.00 x 450 x 300
= 926.44 kN
Puy1 = Puc + Pus(Total)
= 926.44 + (340.98)
= 1267.42 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 286/300
= 0.397
Muc = 926.44 x (0.5 x 300 - 0.397 x 300) = 28.77 kN-m
Muy1 = Muc + Mus(Total)
= 28.77 + (21.69)
= 50.46 kN-m
Pu/Puz = 0.743
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.743, an = 1.906
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((49.63/71.30)1.906) + ((33.41/50.46)1.906)
= 0.957 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG10 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 8 nos. (1608 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1732.79 kN
MomentX,(Mx) = 38.87 kN-m
MomentY,(My) = 34.58 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C180.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 35.869 kN.m
My_MinEccen = 34.656 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 38.873 kN.m
My = max(My,My_MinEccen) + MuaddY = 34.656 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 38.87/1732.79 = 22 mm
Actual eccenY = My / P = 34.58/1732.79 = 20 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(22,21) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(20,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 16 nos. (3217 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00311 354.90 8.92 345.98 139.13 177 24.63
2 402 0.00271 350.48 8.92 341.56 137.35 126 17.36
3 402 0.00230 338.85 8.92 329.93 132.67 76 10.06
4 402 0.00190 323.34 8.90 314.44 126.44 25 3.20
5 402 0.00149 293.49 8.34 285.15 114.66 -25 -2.90
6 402 0.00108 216.90 7.05 209.85 84.39 -76 -6.40
7 402 0.00068 135.70 5.03 130.67 52.55 -126 -6.64
8 402 0.00027 54.49 2.26 52.23 21.00 -177 -3.72
Total808.20 35.59
xux = 436 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 436/450 = 0.349
Puc = 0.349 x 20.00 x 300 x 450
= 941.63 kN
Pux1 = Puc + Pus(Total)
= 941.63 + (808.20)
= 1749.82 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 436/450
= 0.403
Muc = 941.63 x (0.5 x 450 - 0.403 x 450) = 41.10 kN-m
Mux1 = Muc + Mus(Total)
= 41.10 + (35.59)
= 76.69 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1608 0.00289 352.95 8.92 344.03 553.37 102 56.44
2 1608 0.00063 126.93 4.76 122.17 196.51 -102 -20.04
Total749.88 36.40
xuy = 309 mm               Puc = C1.fck.B.D
ku = 1.031
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.027
C1 = 0.446 x (1 - C3/6) = 0.370
Puc = 0.370 x 20.00 x 450 x 300
= 998.00 kN
Puy1 = Puc + Pus(Total)
= 998.00 + (749.88)
= 1747.88 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.426
Muc = 998.00 x (0.5 x 300 - 0.426 x 300) = 22.09 kN-m
Muy1 = Muc + Mus(Total)
= 22.09 + (36.40)
= 58.49 kN-m
Pu/Puz = 0.792
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.792, an = 1.987
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((38.87/76.69)1.987) + ((34.66/58.49)1.987)
= 0.613 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG10 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 16 nos. (3217 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 2203.72 kN
MomentX,(Mx) = 20.93 kN-m
MomentY,(My) = 22.73 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C180.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 47.931 kN.m
My_MinEccen = 44.074 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 47.931 kN.m
My = max(My,My_MinEccen) + MuaddY = 44.074 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 20.93/2203.72 = 9 mm
Actual eccenY = My / P = 22.73/2203.72 = 10 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(9,22) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(10,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 16 nos. (5027 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00310 354.76 8.92 345.84 217.30 175 38.03
2 628 0.00270 350.18 8.92 341.26 214.42 125 26.80
3 628 0.00230 338.60 8.92 329.68 207.14 75 15.54
4 628 0.00189 323.19 8.90 314.30 197.48 25 4.94
5 628 0.00149 293.71 8.35 285.36 179.30 -25 -4.48
6 628 0.00109 218.28 7.08 211.20 132.70 -75 -9.95
7 628 0.00069 137.99 5.09 132.90 83.50 -125 -10.44
8 628 0.00029 57.71 2.39 55.32 34.76 -175 -6.08
Total1266.60 54.35
xux = 436 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 436/450 = 0.349
Puc = 0.349 x 20.00 x 300 x 450
= 941.63 kN
Pux1 = Puc + Pus(Total)
= 941.63 + (1266.60)
= 2208.23 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 436/450
= 0.403
Muc = 941.63 x (0.5 x 450 - 0.403 x 450) = 41.10 kN-m
Mux1 = Muc + Mus(Total)
= 41.10 + (54.35)
= 95.45 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 2513 0.00285 352.56 8.92 343.64 863.67 100 86.37
2 2513 0.00069 138.15 5.10 133.05 334.39 -100 -33.44
Total1198.06 52.93
xuy = 314 mm               Puc = C1.fck.B.D
ku = 1.047
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.976
C1 = 0.446 x (1 - C3/6) = 0.373
Puc = 0.373 x 20.00 x 450 x 300
= 1008.29 kN
Puy1 = Puc + Pus(Total)
= 1008.29 + (1198.06)
= 2206.35 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.431
Muc = 1008.29 x (0.5 x 300 - 0.431 x 300) = 20.99 kN-m
Muy1 = Muc + Mus(Total)
= 20.99 + (52.93)
= 73.92 kN-m
Pu/Puz = 0.806
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.806, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((47.93/95.45)2.000) + ((44.07/73.92)2.000)
= 0.608 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 20 = 320 mm
               iii) 300 mm
               Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG10 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 16 nos. (5027 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG11 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 558.65 kN
MomentX,(Mx) = 34.86 kN-m
MomentY,(My) = 56.09 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C190.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 11.564 kN.m
My_MinEccen = 11.173 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 34.862 kN.m
My = max(My,My_MinEccen) + MuaddY = 56.087 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 34.86/558.65 = 62 mm
Actual eccenY = My / P = 56.09/558.65 = 100 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(62,21) = 62 mm
eccenY = max(Actual eccenY,eccenYMin) = max(100,20) = 100 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(62 mm) > 0.05 x 300(15 mm)
and eccenY(100 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00284 352.51 8.92 343.59 77.72 179 13.91
2 226 0.00156 299.98 8.49 291.49 65.93 90 5.90
3 226 0.00028 55.40 2.30 53.10 12.01 0 0.00
4 226 -0.00101 -201.01 0.00 -201.01 -45.47 -90 4.07
5 226 -0.00229 -338.29 0.00 -338.29 -76.52 -179 13.70
Total33.67 37.58
xux = 244 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 244/450 = 0.195
Puc = 0.195 x 20.00 x 300 x 450
= 527.77 kN
Pux1 = Puc + Pus(Total)
= 527.77 + (33.67)
= 561.44 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 244/450
= 0.226
Muc = 527.77 x (0.5 x 450 - 0.226 x 450) = 65.10 kN-m
Mux1 = Muc + Mus(Total)
= 65.10 + (37.58)
= 102.68 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00255 346.30 8.92 337.38 190.78 104 19.84
2 565 -0.00177 -315.32 0.00 -315.32 -178.31 -104 18.54
Total12.47 38.39
xuy = 169 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 169/300 = 0.203
Puc = 0.203 x 20.00 x 450 x 300
= 546.75 kN
Puy1 = Puc + Pus(Total)
= 546.75 + (12.47)
= 559.22 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 169/300
= 0.234
Muc = 546.75 x (0.5 x 300 - 0.234 x 300) = 43.63 kN-m
Muy1 = Muc + Mus(Total)
= 43.63 + (38.39)
= 82.02 kN-m
Pu/Puz = 0.359
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.359, an = 1.265
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((34.86/102.68)1.265) + ((56.09/82.02)1.265)
= 0.873 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG11 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1113.06 kN
MomentX,(Mx) = 36.77 kN-m
MomentY,(My) = 51.38 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C190.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 23.040 kN.m
My_MinEccen = 22.261 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 36.770 kN.m
My = max(My,My_MinEccen) + MuaddY = 51.377 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 36.77/1113.06 = 33 mm
Actual eccenY = My / P = 51.38/1113.06 = 46 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(33,21) = 33 mm
eccenY = max(Actual eccenY,eccenYMin) = max(46,20) = 46 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(33 mm) > 0.05 x 300(15 mm)
and eccenY(46 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 6 nos. + #16 - 4 nos. (1483 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00306 354.44 8.92 345.52 138.94 177 24.59
2 226 0.00224 336.71 8.92 327.79 74.14 88 6.49
3 226 0.00144 289.17 8.23 280.94 63.55 0 0.00
4 226 0.00065 129.15 4.83 124.32 28.12 -88 -2.46
5 402 -0.00017 -34.34 0.00 -34.34 -13.81 -177 2.44
Total290.95 31.06
xux = 383 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 383/450 = 0.307
Puc = 0.307 x 20.00 x 300 x 450
= 827.72 kN
Pux1 = Puc + Pus(Total)
= 827.72 + (290.95)
= 1118.66 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 383/450
= 0.354
Muc = 827.72 x (0.5 x 450 - 0.354 x 450) = 54.29 kN-m
Mux1 = Muc + Mus(Total)
= 54.29 + (31.06)
= 85.35 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 741 0.00285 352.62 8.92 343.70 254.83 102 25.99
2 741 0.00011 21.95 0.95 20.99 15.57 -102 -1.59
Total270.39 24.40
xuy = 260 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 260/300 = 0.312
Puc = 0.312 x 20.00 x 450 x 300
= 842.91 kN
Puy1 = Puc + Pus(Total)
= 842.91 + (270.39)
= 1113.30 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 260/300
= 0.361
Muc = 842.91 x (0.5 x 300 - 0.361 x 300) = 35.21 kN-m
Muy1 = Muc + Mus(Total)
= 35.21 + (24.40)
= 59.62 kN-m
Pu/Puz = 0.669
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.669, an = 1.782
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((36.77/85.35)1.782) + ((51.38/59.62)1.782)
= 0.990 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG11 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 6 nos. + #16 - 4 nos. (1483 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1685.34 kN
MomentX,(Mx) = 22.68 kN-m
MomentY,(My) = 54.55 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C190.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 34.887 kN.m
My_MinEccen = 33.707 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 34.887 kN.m
My = max(My,My_MinEccen) + MuaddY = 54.547 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 22.68/1685.34 = 13 mm
Actual eccenY = My / P = 54.55/1685.34 = 32 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(13,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(32,20) = 32 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(32 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 16 nos. (3217 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00310 354.79 8.92 345.87 139.08 177 24.62
2 402 0.00268 349.80 8.92 340.88 137.08 126 17.33
3 402 0.00226 337.39 8.92 328.47 132.08 76 10.02
4 402 0.00184 320.00 8.87 311.14 125.12 25 3.16
5 402 0.00142 284.71 8.18 276.53 111.20 -25 -2.81
6 402 0.00100 200.80 6.71 194.09 78.05 -76 -5.92
7 402 0.00058 116.89 4.45 112.44 45.21 -126 -5.72
8 402 0.00016 32.98 1.41 31.57 12.69 -177 -2.25
Total780.52 38.44
xux = 422 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 422/450 = 0.338
Puc = 0.338 x 20.00 x 300 x 450
= 911.25 kN
Pux1 = Puc + Pus(Total)
= 911.25 + (780.52)
= 1691.77 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 422/450
= 0.390
Muc = 911.25 x (0.5 x 450 - 0.390 x 450) = 45.11 kN-m
Mux1 = Muc + Mus(Total)
= 45.11 + (38.44)
= 83.54 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1608 0.00294 353.38 8.92 344.45 554.05 102 56.51
2 1608 0.00056 112.00 4.30 107.70 173.24 -102 -17.67
Total727.30 38.84
xuy = 300 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 300/300 = 0.360
Puc = 0.360 x 20.00 x 450 x 300
= 972.00 kN
Puy1 = Puc + Pus(Total)
= 972.00 + (727.30)
= 1699.30 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 300/300
= 0.416
Muc = 972.00 x (0.5 x 300 - 0.416 x 300) = 24.49 kN-m
Muy1 = Muc + Mus(Total)
= 24.49 + (38.84)
= 63.34 kN-m
Pu/Puz = 0.770
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.770, an = 1.951
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((34.89/83.54)1.951) + ((54.55/63.34)1.951)
= 0.929 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG11 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 16 nos. (3217 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 2226.13 kN
MomentX,(Mx) = 10.01 kN-m
MomentY,(My) = 59.82 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C190.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 46.081 kN.m
My_MinEccen = 44.523 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 46.081 kN.m
My = max(My,My_MinEccen) + MuaddY = 59.817 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 10.01/2226.13 = 4 mm
Actual eccenY = My / P = 59.82/2226.13 = 27 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(4,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(27,20) = 27 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(27 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 16 nos. (5027 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00310 354.82 8.92 345.90 217.33 175 38.03
2 628 0.00271 350.51 8.92 341.59 214.63 125 26.83
3 628 0.00231 339.30 8.92 330.38 207.59 75 15.57
4 628 0.00192 324.78 8.91 315.88 198.47 25 4.96
5 628 0.00152 296.72 8.42 288.31 181.15 -25 -4.53
6 628 0.00113 225.93 7.23 218.70 137.41 -75 -10.31
7 628 0.00073 146.91 5.35 141.56 88.95 -125 -11.12
8 628 0.00034 67.90 2.77 65.13 40.92 -175 -7.16
Total1286.45 52.28
xux = 443 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 443/450 = 0.354
Puc = 0.354 x 20.00 x 300 x 450
= 956.81 kN
Pux1 = Puc + Pus(Total)
= 956.81 + (1286.45)
= 2243.26 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 443/450
= 0.410
Muc = 956.81 x (0.5 x 450 - 0.410 x 450) = 38.97 kN-m
Mux1 = Muc + Mus(Total)
= 38.97 + (52.28)
= 91.24 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 2513 0.00283 352.38 8.92 343.46 863.21 100 86.32
2 2513 0.00072 144.60 5.28 139.32 350.14 -100 -35.01
Total1213.35 51.31
xuy = 319 mm               Puc = C1.fck.B.D
ku = 1.063
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.929
C1 = 0.446 x (1 - C3/6) = 0.377
Puc = 0.377 x 20.00 x 450 x 300
= 1017.83 kN
Puy1 = Puc + Pus(Total)
= 1017.83 + (1213.35)
= 2231.18 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.435
Muc = 1017.83 x (0.5 x 300 - 0.435 x 300) = 19.97 kN-m
Muy1 = Muc + Mus(Total)
= 19.97 + (51.31)
= 71.28 kN-m
Pu/Puz = 0.814
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.814, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((46.08/91.24)2.000) + ((59.82/71.28)2.000)
= 0.959 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 20 = 320 mm
               iii) 300 mm
               Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG11 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 16 nos. (5027 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 2853.12 kN
MomentX,(Mx) = 14.31 kN-m
MomentY,(My) = 41.16 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C190.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 62.055 kN.m
My_MinEccen = 57.062 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 62.055 kN.m
My = max(My,My_MinEccen) + MuaddY = 57.062 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 14.31/2853.12 = 5 mm
Actual eccenY = My / P = 41.16/2853.12 = 14 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(5,22) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(14,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #32 - 10 nos. (8042 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1608 0.00304 354.24 8.92 345.32 555.45 169 93.87
2 1608 0.00234 340.38 8.92 331.46 533.14 85 45.05
3 1608 0.00165 307.87 8.64 299.23 481.30 0 0.00
4 1608 0.00095 190.70 6.48 184.23 296.33 -85 -25.04
5 1608 0.00026 51.65 2.16 49.50 79.62 -169 -13.46
Total1945.85 100.43
xux = 425 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 425/450 = 0.340
Puc = 0.340 x 20.00 x 300 x 450
= 918.84 kN
Pux1 = Puc + Pus(Total)
= 918.84 + (1945.85)
= 2864.69 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 425/450
= 0.393
Muc = 918.84 x (0.5 x 450 - 0.393 x 450) = 44.14 kN-m
Mux1 = Muc + Mus(Total)
= 44.14 + (100.43)
= 144.57 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 4021 0.00285 352.56 8.92 343.64 1381.85 94 129.89
2 4021 0.00065 130.67 4.88 125.79 505.83 -94 -47.55
Total1887.69 82.35
xuy = 300 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 300/300 = 0.360
Puc = 0.360 x 20.00 x 450 x 300
= 972.00 kN
Puy1 = Puc + Pus(Total)
= 972.00 + (1887.69)
= 2859.69 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 300/300
= 0.416
Muc = 972.00 x (0.5 x 300 - 0.416 x 300) = 24.49 kN-m
Muy1 = Muc + Mus(Total)
= 24.49 + (82.35)
= 106.84 kN-m
Pu/Puz = 0.783
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.783, an = 1.971
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((62.06/144.57)1.971) + ((57.06/106.84)1.971)
= 0.479 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 32/4 = 8 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (8, 6) = 8 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 32 = 512 mm
               iii) 300 mm
               Required spacing = minimum of (300, 512, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG11 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #32 - 10 nos. (8042 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG12 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 386.62 kN
MomentX,(Mx) = 7.65 kN-m
MomentY,(My) = 85.25 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C200.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 8.003 kN.m
My_MinEccen = 7.732 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 8.003 kN.m
My = max(My,My_MinEccen) + MuaddY = 85.251 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 7.65/386.62 = 20 mm
Actual eccenY = My / P = 85.25/386.62 = 221 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(20,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(221,20) = 221 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(221 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 6 nos. + #16 - 4 nos. (1483 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00267 349.46 8.92 340.54 136.94 177 24.24
2 226 0.00112 223.86 7.19 216.67 49.01 88 4.29
3 226 -0.00040 -79.13 0.00 -79.13 -17.90 0 0.00
4 226 -0.00191 -324.22 0.00 -324.22 -73.34 -88 6.42
5 402 -0.00346 -357.93 0.00 -357.93 -143.93 -177 25.48
Total-49.22 60.42
xux = 202 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 202/450 = 0.162
Puc = 0.162 x 20.00 x 300 x 450
= 436.64 kN
Pux1 = Puc + Pus(Total)
= 436.64 + (-49.22)
= 387.42 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 202/450
= 0.187
Muc = 436.64 x (0.5 x 450 - 0.187 x 450) = 61.53 kN-m
Mux1 = Muc + Mus(Total)
= 61.53 + (60.42)
= 121.94 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 741 0.00218 334.53 8.92 325.61 241.41 102 24.62
2 741 -0.00340 -357.44 0.00 -357.44 -265.01 -102 27.03
Total-23.60 51.66
xuy = 128 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 128/300 = 0.153
Puc = 0.153 x 20.00 x 450 x 300
= 413.86 kN
Puy1 = Puc + Pus(Total)
= 413.86 + (-23.60)
= 390.25 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 128/300
= 0.177
Muc = 413.86 x (0.5 x 300 - 0.177 x 300) = 40.09 kN-m
Muy1 = Muc + Mus(Total)
= 40.09 + (51.66)
= 91.74 kN-m
Pu/Puz = 0.232
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.232, an = 1.054
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((8.00/121.94)1.054) + ((85.25/91.74)1.054)
= 0.982 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG12 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 6 nos. + #16 - 4 nos. (1483 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 781.33 kN
MomentX,(Mx) = 7.26 kN-m
MomentY,(My) = 77.98 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C200.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 16.174 kN.m
My_MinEccen = 15.627 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 16.174 kN.m
My = max(My,My_MinEccen) + MuaddY = 77.977 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 7.26/781.33 = 9 mm
Actual eccenY = My / P = 77.98/781.33 = 100 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(9,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(100,20) = 100 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(100 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 14 nos. (1583 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00294 353.42 8.92 344.50 77.92 179 13.95
2 226 0.00222 336.00 8.92 327.08 73.98 119 8.83
3 226 0.00150 294.85 8.37 286.47 64.80 60 3.87
4 226 0.00078 156.97 5.63 151.34 34.23 -0 -0.00
5 226 0.00006 12.97 0.57 12.40 2.80 -60 -0.17
6 226 -0.00066 -131.04 0.00 -131.04 -29.64 -119 3.54
7 226 -0.00138 -275.04 0.00 -275.04 -62.21 -179 11.14
Total161.89 41.15
xux = 290 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 290/450 = 0.232
Puc = 0.232 x 20.00 x 300 x 450
= 626.48 kN
Pux1 = Puc + Pus(Total)
= 626.48 + (161.89)
= 788.37 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 290/450
= 0.268
Muc = 626.48 x (0.5 x 450 - 0.268 x 450) = 65.37 kN-m
Mux1 = Muc + Mus(Total)
= 65.37 + (41.15)
= 106.52 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 792 0.00271 350.41 8.92 341.49 270.35 104 28.12
2 792 -0.00089 -177.01 0.00 -177.01 -140.14 -104 14.57
Total130.21 42.69
xuy = 203 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 203/300 = 0.243
Puc = 0.243 x 20.00 x 450 x 300
= 656.86 kN
Puy1 = Puc + Pus(Total)
= 656.86 + (130.21)
= 787.07 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 203/300
= 0.281
Muc = 656.86 x (0.5 x 300 - 0.281 x 300) = 43.13 kN-m
Muy1 = Muc + Mus(Total)
= 43.13 + (42.69)
= 85.82 kN-m
Pu/Puz = 0.461
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.461, an = 1.436
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((16.17/106.52)1.436) + ((77.98/85.82)1.436)
= 0.938 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG12 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 14 nos. (1583 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1185.41 kN
MomentX,(Mx) = 5.82 kN-m
MomentY,(My) = 77.19 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C200.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 24.538 kN.m
My_MinEccen = 23.708 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 24.538 kN.m
My = max(My,My_MinEccen) + MuaddY = 77.191 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 5.82/1185.41 = 5 mm
Actual eccenY = My / P = 77.19/1185.41 = 65 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(5,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(65,20) = 65 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(65 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 14 nos. (2815 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00300 353.90 8.92 344.98 138.72 177 24.55
2 402 0.00238 341.87 8.92 332.95 133.89 118 15.80
3 402 0.00177 315.41 8.80 306.61 123.29 59 7.27
4 402 0.00115 230.89 7.33 223.56 89.90 0 0.00
5 402 0.00054 107.88 4.16 103.72 41.71 -59 -2.46
6 402 -0.00008 -15.13 0.00 -15.13 -6.08 -118 0.72
7 402 -0.00069 -138.14 0.00 -138.14 -55.55 -177 9.83
Total465.87 55.72
xux = 336 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 336/450 = 0.269
Puc = 0.269 x 20.00 x 300 x 450
= 725.20 kN
Pux1 = Puc + Pus(Total)
= 725.20 + (465.87)
= 1191.08 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 336/450
= 0.310
Muc = 725.20 x (0.5 x 450 - 0.310 x 450) = 61.88 kN-m
Mux1 = Muc + Mus(Total)
= 61.88 + (55.72)
= 117.60 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1407 0.00279 352.07 8.92 343.15 482.95 102 49.26
2 1407 -0.00023 -45.19 0.00 -45.19 -63.60 -102 6.49
Total419.35 55.75
xuy = 237 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 237/300 = 0.284
Puc = 0.284 x 20.00 x 450 x 300
= 766.97 kN
Puy1 = Puc + Pus(Total)
= 766.97 + (419.35)
= 1186.32 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 237/300
= 0.328
Muc = 766.97 x (0.5 x 300 - 0.328 x 300) = 39.52 kN-m
Muy1 = Muc + Mus(Total)
= 39.52 + (55.75)
= 95.27 kN-m
Pu/Puz = 0.574
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.574, an = 1.623
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((24.54/117.60)1.623) + ((77.19/95.27)1.623)
= 0.789 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG12 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 14 nos. (2815 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1604.80 kN
MomentX,(Mx) = 3.94 kN-m
MomentY,(My) = 76.38 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C200.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 33.219 kN.m
My_MinEccen = 32.096 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 33.219 kN.m
My = max(My,My_MinEccen) + MuaddY = 76.379 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 3.94/1604.80 = 2 mm
Actual eccenY = My / P = 76.38/1604.80 = 48 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(2,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(48,20) = 48 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(48 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 12 nos. (3770 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00304 354.26 8.92 345.34 216.98 175 37.97
2 628 0.00240 342.39 8.92 333.47 209.53 105 22.00
3 628 0.00176 314.60 8.79 305.82 192.15 35 6.73
4 628 0.00111 222.87 7.17 215.70 135.53 -35 -4.74
5 628 0.00047 94.41 3.71 90.69 56.99 -105 -5.98
6 628 -0.00017 -34.05 0.00 -34.05 -21.39 -175 3.74
Total789.78 59.71
xux = 381 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 381/450 = 0.305
Puc = 0.305 x 20.00 x 300 x 450
= 823.92 kN
Pux1 = Puc + Pus(Total)
= 823.92 + (789.78)
= 1613.70 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 381/450
= 0.353
Muc = 823.92 x (0.5 x 450 - 0.353 x 450) = 54.64 kN-m
Mux1 = Muc + Mus(Total)
= 54.64 + (59.71)
= 114.36 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1885 0.00285 352.59 8.92 343.67 647.81 100 64.78
2 1885 0.00025 50.72 2.12 48.61 91.62 -100 -9.16
Total739.43 55.62
xuy = 270 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 270/300 = 0.323
Puc = 0.323 x 20.00 x 450 x 300
= 873.28 kN
Puy1 = Puc + Pus(Total)
= 873.28 + (739.43)
= 1612.71 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 270/300
= 0.374
Muc = 873.28 x (0.5 x 300 - 0.374 x 300) = 33.08 kN-m
Muy1 = Muc + Mus(Total)
= 33.08 + (55.62)
= 88.69 kN-m
Pu/Puz = 0.682
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.682, an = 1.803
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((33.22/114.36)1.803) + ((76.38/88.69)1.803)
= 0.871 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 20 = 320 mm
               iii) 300 mm
               Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG12 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 12 nos. (3770 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 2044.35 kN
MomentX,(Mx) = 1.14 kN-m
MomentY,(My) = 51.67 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C200.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 44.465 kN.m
My_MinEccen = 40.887 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 44.465 kN.m
My = max(My,My_MinEccen) + MuaddY = 51.666 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 1.14/2044.35 = 1 mm
Actual eccenY = My / P = 51.67/2044.35 = 25 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(1,22) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(25,20) = 25 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(25 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 14 nos. (4398 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00310 354.79 8.92 345.87 217.32 175 38.03
2 628 0.00264 348.64 8.92 339.72 213.45 117 24.90
3 628 0.00217 334.08 8.92 325.16 204.30 58 11.92
4 628 0.00171 311.57 8.73 302.84 190.28 0 0.00
5 628 0.00124 248.68 7.64 241.04 151.45 -58 -8.83
6 628 0.00078 155.76 5.59 150.17 94.35 -117 -11.01
7 628 0.00031 62.84 2.58 60.26 37.86 -175 -6.63
Total1109.02 48.38
xux = 439 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 439/450 = 0.352
Puc = 0.352 x 20.00 x 300 x 450
= 949.22 kN
Pux1 = Puc + Pus(Total)
= 949.22 + (1109.02)
= 2058.24 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 439/450
= 0.406
Muc = 949.22 x (0.5 x 450 - 0.406 x 450) = 40.05 kN-m
Mux1 = Muc + Mus(Total)
= 40.05 + (48.38)
= 88.43 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 2199 0.00285 352.56 8.92 343.64 755.71 100 75.57
2 2199 0.00069 138.15 5.10 133.05 292.59 -100 -29.26
Total1048.30 46.31
xuy = 314 mm               Puc = C1.fck.B.D
ku = 1.047
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.976
C1 = 0.446 x (1 - C3/6) = 0.373
Puc = 0.373 x 20.00 x 450 x 300
= 1008.29 kN
Puy1 = Puc + Pus(Total)
= 1008.29 + (1048.30)
= 2056.59 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.431
Muc = 1008.29 x (0.5 x 300 - 0.431 x 300) = 20.99 kN-m
Muy1 = Muc + Mus(Total)
= 20.99 + (46.31)
= 67.30 kN-m
Pu/Puz = 0.803
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.803, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((44.46/88.43)2.000) + ((51.67/67.30)2.000)
= 0.842 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 20 = 320 mm
               iii) 300 mm
               Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG12 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 14 nos. (4398 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG13 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 283.57 kN
MomentX,(Mx) = 18.07 kN-m
MomentY,(My) = 44.39 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C210.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 5.870 kN.m
My_MinEccen = 5.671 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 18.070 kN.m
My = max(My,My_MinEccen) + MuaddY = 44.386 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 18.07/283.57 = 64 mm
Actual eccenY = My / P = 44.39/283.57 = 157 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(64,21) = 64 mm
eccenY = max(Actual eccenY,eccenYMin) = max(157,20) = 157 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(64 mm) > 0.05 x 300(15 mm)
and eccenY(157 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00258 347.28 8.92 338.36 76.53 179 13.70
2 226 0.00080 160.41 5.72 154.69 34.99 90 3.13
3 226 -0.00098 -196.00 0.00 -196.00 -44.33 0 0.00
4 226 -0.00276 -351.82 0.00 -351.82 -79.58 -90 7.12
5 226 -0.00454 -360.90 0.00 -360.90 -81.63 -179 14.61
Total-94.02 38.57
xux = 176 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 176/450 = 0.141
Puc = 0.141 x 20.00 x 300 x 450
= 379.69 kN
Pux1 = Puc + Pus(Total)
= 379.69 + (-94.02)
= 285.66 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 176/450
= 0.163
Muc = 379.69 x (0.5 x 450 - 0.163 x 450) = 57.67 kN-m
Mux1 = Muc + Mus(Total)
= 57.67 + (38.57)
= 96.23 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00183 319.48 8.86 310.62 175.65 104 18.27
2 565 -0.00570 -360.90 0.00 -360.90 -204.08 -104 21.22
Total-28.43 39.49
xuy = 97 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 97/300 = 0.116
Puc = 0.116 x 20.00 x 450 x 300
= 313.24 kN
Puy1 = Puc + Pus(Total)
= 313.24 + (-28.43)
= 284.81 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 97/300
= 0.134
Muc = 313.24 x (0.5 x 300 - 0.134 x 300) = 34.39 kN-m
Muy1 = Muc + Mus(Total)
= 34.39 + (39.49)
= 73.88 kN-m
Pu/Puz = 0.182
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.182, an = 1.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((18.07/96.23)1.000) + ((44.39/73.88)1.000)
= 0.789 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG13 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 586.34 kN
MomentX,(Mx) = 16.25 kN-m
MomentY,(My) = 40.17 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C210.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 12.137 kN.m
My_MinEccen = 11.727 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 16.248 kN.m
My = max(My,My_MinEccen) + MuaddY = 40.171 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 16.25/586.34 = 28 mm
Actual eccenY = My / P = 40.17/586.34 = 69 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(28,21) = 28 mm
eccenY = max(Actual eccenY,eccenYMin) = max(69,20) = 69 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(28 mm) > 0.05 x 300(15 mm)
and eccenY(69 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00286 352.67 8.92 343.75 77.75 179 13.92
2 226 0.00161 305.12 8.59 296.53 67.07 90 6.00
3 226 0.00037 73.43 2.97 70.45 15.94 0 0.00
4 226 -0.00088 -175.81 0.00 -175.81 -39.77 -90 3.56
5 226 -0.00213 -332.34 0.00 -332.34 -75.17 -179 13.46
Total45.82 36.94
xux = 251 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 251/450 = 0.201
Puc = 0.201 x 20.00 x 300 x 450
= 542.95 kN
Pux1 = Puc + Pus(Total)
= 542.95 + (45.82)
= 588.78 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 251/450
= 0.232
Muc = 542.95 x (0.5 x 450 - 0.232 x 450) = 65.39 kN-m
Mux1 = Muc + Mus(Total)
= 65.39 + (36.94)
= 102.32 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00258 347.28 8.92 338.36 191.34 104 19.90
2 565 -0.00156 -299.82 0.00 -299.82 -169.55 -104 17.63
Total21.79 37.53
xuy = 176 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 176/300 = 0.211
Puc = 0.211 x 20.00 x 450 x 300
= 569.53 kN
Puy1 = Puc + Pus(Total)
= 569.53 + (21.79)
= 591.32 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 176/300
= 0.244
Muc = 569.53 x (0.5 x 300 - 0.244 x 300) = 43.78 kN-m
Muy1 = Muc + Mus(Total)
= 43.78 + (37.53)
= 81.31 kN-m
Pu/Puz = 0.377
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.377, an = 1.294
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((16.25/102.32)1.294) + ((40.17/81.31)1.294)
= 0.494 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG13 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 877.93 kN
MomentX,(Mx) = 14.32 kN-m
MomentY,(My) = 35.65 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C210.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 18.173 kN.m
My_MinEccen = 17.559 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 18.173 kN.m
My = max(My,My_MinEccen) + MuaddY = 35.655 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 14.32/877.93 = 16 mm
Actual eccenY = My / P = 35.65/877.93 = 41 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(16,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(41,20) = 41 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(41 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00301 354.01 8.92 345.09 78.06 179 13.97
2 226 0.00206 330.12 8.92 321.20 72.65 90 6.50
3 226 0.00112 223.40 7.18 216.22 48.91 0 0.00
4 226 0.00017 33.83 1.44 32.38 7.32 -90 -0.66
5 226 -0.00078 -155.75 0.00 -155.75 -35.23 -179 6.31
Total171.71 26.13
xux = 330 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 330/450 = 0.264
Puc = 0.264 x 20.00 x 300 x 450
= 713.81 kN
Pux1 = Puc + Pus(Total)
= 713.81 + (171.71)
= 885.53 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 330/450
= 0.306
Muc = 713.81 x (0.5 x 450 - 0.306 x 450) = 62.48 kN-m
Mux1 = Muc + Mus(Total)
= 62.48 + (26.13)
= 88.60 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00279 352.08 8.92 343.16 194.05 104 20.18
2 565 -0.00041 -82.08 0.00 -82.08 -46.41 -104 4.83
Total147.64 25.01
xuy = 227 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 227/300 = 0.273
Puc = 0.273 x 20.00 x 450 x 300
= 736.59 kN
Puy1 = Puc + Pus(Total)
= 736.59 + (147.64)
= 884.23 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 227/300
= 0.315
Muc = 736.59 x (0.5 x 300 - 0.315 x 300) = 40.83 kN-m
Muy1 = Muc + Mus(Total)
= 40.83 + (25.01)
= 65.83 kN-m
Pu/Puz = 0.564
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.564, an = 1.607
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((18.17/88.60)1.607) + ((35.65/65.83)1.607)
= 0.452 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG13 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1150.83 kN
MomentX,(Mx) = 11.38 kN-m
MomentY,(My) = 28.95 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C210.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 23.822 kN.m
My_MinEccen = 23.017 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 23.822 kN.m
My = max(My,My_MinEccen) + MuaddY = 28.947 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 11.38/1150.83 = 10 mm
Actual eccenY = My / P = 28.95/1150.83 = 25 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(10,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(25,20) = 25 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(25 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00311 354.88 8.92 345.96 78.25 179 14.01
2 226 0.00236 340.85 8.92 331.93 75.08 90 6.72
3 226 0.00160 304.02 8.57 295.45 66.83 0 0.00
4 226 0.00085 169.32 5.95 163.36 36.95 -90 -3.31
5 226 0.00009 18.30 0.80 17.50 3.96 -179 -0.71
Total261.07 16.71
xux = 415 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 415/450 = 0.332
Puc = 0.332 x 20.00 x 300 x 450
= 896.06 kN
Pux1 = Puc + Pus(Total)
= 896.06 + (261.07)
= 1157.14 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 415/450
= 0.384
Muc = 896.06 x (0.5 x 450 - 0.384 x 450) = 46.98 kN-m
Mux1 = Muc + Mus(Total)
= 46.98 + (16.71)
= 63.69 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00293 353.31 8.92 344.39 194.75 104 20.25
2 565 0.00037 73.05 2.96 70.09 39.63 -104 -4.12
Total234.38 16.13
xuy = 284 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 284/300 = 0.340
Puc = 0.340 x 20.00 x 450 x 300
= 918.84 kN
Puy1 = Puc + Pus(Total)
= 918.84 + (234.38)
= 1153.22 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 284/300
= 0.393
Muc = 918.84 x (0.5 x 300 - 0.393 x 300) = 29.43 kN-m
Muy1 = Muc + Mus(Total)
= 29.43 + (16.13)
= 45.56 kN-m
Pu/Puz = 0.739
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.739, an = 1.899
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((23.82/63.69)1.899) + ((28.95/45.56)1.899)
= 0.577 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG13 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 1390.20 kN
MomentX,(Mx) = 5.61 kN-m
MomentY,(My) = 15.81 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C210.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 30.237 kN.m
My_MinEccen = 27.804 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 30.237 kN.m
My = max(My,My_MinEccen) + MuaddY = 27.804 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 5.61/1390.20 = 4 mm
Actual eccenY = My / P = 15.81/1390.20 = 11 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(4,22) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(11,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 14 nos. (1583 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00311 354.88 8.92 345.96 78.25 179 14.01
2 226 0.00266 349.23 8.92 340.31 76.98 119 9.19
3 226 0.00221 335.39 8.92 326.47 73.85 60 4.41
4 226 0.00176 314.60 8.79 305.82 69.17 -0 -0.00
5 226 0.00130 260.99 7.84 253.14 57.26 -60 -3.42
6 226 0.00085 170.64 5.99 164.65 37.24 -119 -4.44
7 226 0.00040 80.30 3.22 77.08 17.43 -179 -3.12
Total410.19 16.62
xux = 457 mm               Puc = C1.fck.B.D
ku = 1.016
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.083
C1 = 0.446 x (1 - C3/6) = 0.366
Puc = 0.366 x 20.00 x 300 x 450
= 986.88 kN
Pux1 = Puc + Pus(Total)
= 986.88 + (410.19)
= 1397.07 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.421
Muc = 986.88 x (0.5 x 450 - 0.421 x 450) = 34.93 kN-m
Mux1 = Muc + Mus(Total)
= 34.93 + (16.62)
= 51.54 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 792 0.00287 352.75 8.92 343.83 272.20 104 28.31
2 792 0.00068 136.19 5.04 131.15 103.83 -104 -10.80
Total376.03 17.51
xuy = 319 mm               Puc = C1.fck.B.D
ku = 1.063
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.929
C1 = 0.446 x (1 - C3/6) = 0.377
Puc = 0.377 x 20.00 x 450 x 300
= 1017.83 kN
Puy1 = Puc + Pus(Total)
= 1017.83 + (376.03)
= 1393.86 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.435
Muc = 1017.83 x (0.5 x 300 - 0.435 x 300) = 19.97 kN-m
Muy1 = Muc + Mus(Total)
= 19.97 + (17.51)
= 37.48 kN-m
Pu/Puz = 0.821
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.821, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((30.24/51.54)2.000) + ((27.80/37.48)2.000)
= 0.894 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG13 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 14 nos. (1583 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG14 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 531.28 kN
MomentX,(Mx) = 39.57 kN-m
MomentY,(My) = 13.59 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C220.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 10.998 kN.m
My_MinEccen = 10.626 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 39.574 kN.m
My = max(My,My_MinEccen) + MuaddY = 13.586 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 39.57/531.28 = 74 mm
Actual eccenY = My / P = 13.59/531.28 = 26 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(74,21) = 74 mm
eccenY = max(Actual eccenY,eccenYMin) = max(26,20) = 26 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(74 mm) > 0.05 x 300(15 mm)
and eccenY(26 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00282 352.34 8.92 343.42 77.68 179 13.90
2 226 0.00150 294.53 8.37 286.16 64.73 90 5.79
3 226 0.00018 36.30 1.55 34.75 7.86 0 0.00
4 226 -0.00114 -227.71 0.00 -227.71 -51.51 -90 4.61
5 226 -0.00246 -344.05 0.00 -344.05 -77.82 -179 13.93
Total20.94 38.24
xux = 237 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 237/450 = 0.190
Puc = 0.190 x 20.00 x 300 x 450
= 512.58 kN
Pux1 = Puc + Pus(Total)
= 512.58 + (20.94)
= 533.52 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 237/450
= 0.219
Muc = 512.58 x (0.5 x 450 - 0.219 x 450) = 64.73 kN-m
Mux1 = Muc + Mus(Total)
= 64.73 + (38.24)
= 102.97 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00251 345.41 8.92 336.49 190.28 104 19.79
2 565 -0.00196 -326.18 0.00 -326.18 -184.45 -104 19.18
Total5.83 38.97
xuy = 163 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 163/300 = 0.195
Puc = 0.195 x 20.00 x 450 x 300
= 527.77 kN
Puy1 = Puc + Pus(Total)
= 527.77 + (5.83)
= 533.60 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 163/300
= 0.226
Muc = 527.77 x (0.5 x 300 - 0.226 x 300) = 43.40 kN-m
Muy1 = Muc + Mus(Total)
= 43.40 + (38.97)
= 82.37 kN-m
Pu/Puz = 0.341
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.341, an = 1.235
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((39.57/102.97)1.235) + ((13.59/82.37)1.235)
= 0.415 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG14 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1049.43 kN
MomentX,(Mx) = 31.95 kN-m
MomentY,(My) = 12.77 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C220.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 21.723 kN.m
My_MinEccen = 20.989 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 31.951 kN.m
My = max(My,My_MinEccen) + MuaddY = 20.989 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 31.95/1049.43 = 30 mm
Actual eccenY = My / P = 12.77/1049.43 = 12 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(30,21) = 30 mm
eccenY = max(Actual eccenY,eccenYMin) = max(12,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(30 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00308 354.56 8.92 345.64 78.18 179 13.99
2 226 0.00225 336.96 8.92 328.04 74.20 90 6.64
3 226 0.00143 285.19 8.19 277.00 62.66 0 0.00
4 226 0.00060 120.18 4.55 115.63 26.15 -90 -2.34
5 226 -0.00022 -44.82 0.00 -44.82 -10.14 -179 1.81
Total231.05 20.11
xux = 380 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 380/450 = 0.304
Puc = 0.304 x 20.00 x 300 x 450
= 820.13 kN
Pux1 = Puc + Pus(Total)
= 820.13 + (231.05)
= 1051.18 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 380/450
= 0.351
Muc = 820.13 x (0.5 x 450 - 0.351 x 450) = 54.99 kN-m
Mux1 = Muc + Mus(Total)
= 54.99 + (20.11)
= 75.10 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00289 352.91 8.92 343.99 194.52 104 20.23
2 565 0.00011 22.67 0.98 21.68 12.26 -104 -1.28
Total206.78 18.95
xuy = 263 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 263/300 = 0.315
Puc = 0.315 x 20.00 x 450 x 300
= 850.50 kN
Puy1 = Puc + Pus(Total)
= 850.50 + (206.78)
= 1057.28 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 263/300
= 0.364
Muc = 850.50 x (0.5 x 300 - 0.364 x 300) = 34.70 kN-m
Muy1 = Muc + Mus(Total)
= 34.70 + (18.95)
= 53.66 kN-m
Pu/Puz = 0.674
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.674, an = 1.790
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((31.95/75.10)1.790) + ((20.99/53.66)1.790)
= 0.403 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG14 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1584.54 kN
MomentX,(Mx) = 34.96 kN-m
MomentY,(My) = 10.12 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C220.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 32.800 kN.m
My_MinEccen = 31.691 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 34.963 kN.m
My = max(My,My_MinEccen) + MuaddY = 31.691 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 34.96/1584.54 = 22 mm
Actual eccenY = My / P = 10.12/1584.54 = 6 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(22,21) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(6,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 20 nos. (2262 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00308 354.63 8.92 345.71 78.20 179 14.00
2 226 0.00279 352.06 8.92 343.14 77.62 139 10.81
3 226 0.00250 345.02 8.92 336.10 76.02 99 7.56
4 226 0.00220 335.19 8.92 326.27 73.80 60 4.40
5 226 0.00191 324.15 8.90 315.25 71.31 20 1.42
6 226 0.00162 305.40 8.59 296.81 67.14 -20 -1.34
7 226 0.00132 264.59 7.90 256.69 58.06 -60 -3.46
8 226 0.00103 205.92 6.82 199.10 45.04 -99 -4.48
9 226 0.00074 147.25 5.36 141.90 32.10 -139 -4.47
10 226 0.00044 88.59 3.51 85.07 19.24 -179 -3.44
Total598.52 20.99
xux = 464 mm               Puc = C1.fck.B.D
ku = 1.031
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.027
C1 = 0.446 x (1 - C3/6) = 0.370
Puc = 0.370 x 20.00 x 300 x 450
= 998.00 kN
Pux1 = Puc + Pus(Total)
= 998.00 + (598.52)
= 1596.52 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.426
Muc = 998.00 x (0.5 x 450 - 0.426 x 450) = 33.14 kN-m
Mux1 = Muc + Mus(Total)
= 33.14 + (20.99)
= 54.13 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1131 0.00283 352.39 8.92 343.47 388.46 104 40.40
2 1131 0.00074 148.58 5.40 143.19 161.94 -104 -16.84
Total550.40 23.56
xuy = 328 mm               Puc = C1.fck.B.D
ku = 1.094
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.843
C1 = 0.446 x (1 - C3/6) = 0.383
Puc = 0.383 x 20.00 x 450 x 300
= 1034.93 kN
Puy1 = Puc + Pus(Total)
= 1034.93 + (550.40)
= 1585.32 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.442
Muc = 1034.93 x (0.5 x 300 - 0.442 x 300) = 18.14 kN-m
Muy1 = Muc + Mus(Total)
= 18.14 + (23.56)
= 41.69 kN-m
Pu/Puz = 0.835
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.835, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((34.96/54.13)2.000) + ((31.69/41.69)2.000)
= 0.995 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG14 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 20 nos. (2262 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 2147.07 kN
MomentX,(Mx) = 42.32 kN-m
MomentY,(My) = 6.52 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C220.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 44.444 kN.m
My_MinEccen = 42.941 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 44.444 kN.m
My = max(My,My_MinEccen) + MuaddY = 42.941 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 42.32/2147.07 = 20 mm
Actual eccenY = My / P = 6.52/2147.07 = 3 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(20,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(3,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 14 nos. (4398 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00305 354.37 8.92 345.45 217.05 175 37.98
2 628 0.00262 348.29 8.92 339.37 213.23 117 24.88
3 628 0.00219 334.83 8.92 325.91 204.78 58 11.95
4 628 0.00176 315.00 8.79 306.20 192.39 0 0.00
5 628 0.00133 266.56 7.93 258.63 162.50 -58 -9.48
6 628 0.00090 180.52 6.23 174.29 109.51 -117 -12.78
7 628 0.00047 94.49 3.72 90.77 57.03 -175 -9.98
Total1156.49 42.57
xux = 464 mm               Puc = C1.fck.B.D
ku = 1.031
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.027
C1 = 0.446 x (1 - C3/6) = 0.370
Puc = 0.370 x 20.00 x 300 x 450
= 998.00 kN
Pux1 = Puc + Pus(Total)
= 998.00 + (1156.49)
= 2154.49 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.426
Muc = 998.00 x (0.5 x 450 - 0.426 x 450) = 33.14 kN-m
Mux1 = Muc + Mus(Total)
= 33.14 + (42.57)
= 75.71 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 2199 0.00275 351.60 8.92 342.68 753.59 100 75.36
2 2199 0.00084 167.52 5.91 161.61 355.41 -100 -35.54
Total1109.00 39.82
xuy = 338 mm               Puc = C1.fck.B.D
ku = 1.125
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.769
C1 = 0.446 x (1 - C3/6) = 0.389
Puc = 0.389 x 20.00 x 450 x 300
= 1049.78 kN
Puy1 = Puc + Pus(Total)
= 1049.78 + (1109.00)
= 2158.77 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.447
Muc = 1049.78 x (0.5 x 300 - 0.447 x 300) = 16.55 kN-m
Muy1 = Muc + Mus(Total)
= 16.55 + (39.82)
= 56.36 kN-m
Pu/Puz = 0.844
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.844, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((44.44/75.71)2.000) + ((42.94/56.36)2.000)
= 0.925 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 20 = 320 mm
               iii) 300 mm
               Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG14 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 14 nos. (4398 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 2759.80 kN
MomentX,(Mx) = 26.29 kN-m
MomentY,(My) = 2.48 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C220.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 60.026 kN.m
My_MinEccen = 55.196 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 60.026 kN.m
My = max(My,My_MinEccen) + MuaddY = 55.196 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 26.29/2759.80 = 10 mm
Actual eccenY = My / P = 2.48/2759.80 = 1 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(10,22) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(1,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #25 - 14 nos. (6872 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 982 0.00306 354.45 8.92 345.53 339.22 173 58.52
2 982 0.00263 348.39 8.92 339.47 333.27 115 38.33
3 982 0.00219 334.79 8.92 325.87 319.92 58 18.40
4 982 0.00176 314.60 8.79 305.82 300.24 0 0.00
5 982 0.00132 264.27 7.89 256.37 251.69 -58 -14.47
6 982 0.00089 177.20 6.15 171.05 167.93 -115 -19.31
7 982 0.00045 90.14 3.57 86.57 84.99 -173 -14.66
Total1797.27 66.79
xux = 457 mm               Puc = C1.fck.B.D
ku = 1.016
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.083
C1 = 0.446 x (1 - C3/6) = 0.366
Puc = 0.366 x 20.00 x 300 x 450
= 986.88 kN
Pux1 = Puc + Pus(Total)
= 986.88 + (1797.27)
= 2784.14 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.421
Muc = 986.88 x (0.5 x 450 - 0.421 x 450) = 34.93 kN-m
Mux1 = Muc + Mus(Total)
= 34.93 + (66.79)
= 101.72 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 3436 0.00274 351.41 8.92 342.49 1176.84 98 114.74
2 3436 0.00084 167.08 5.90 161.19 553.86 -98 -54.00
Total1730.70 60.74
xuy = 333 mm               Puc = C1.fck.B.D
ku = 1.109
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.805
C1 = 0.446 x (1 - C3/6) = 0.386
Puc = 0.386 x 20.00 x 450 x 300
= 1042.61 kN
Puy1 = Puc + Pus(Total)
= 1042.61 + (1730.70)
= 2773.31 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.445
Muc = 1042.61 x (0.5 x 300 - 0.445 x 300) = 17.31 kN-m
Muy1 = Muc + Mus(Total)
= 17.31 + (60.74)
= 78.05 kN-m
Pu/Puz = 0.838
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.838, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((60.03/101.72)2.000) + ((55.20/78.05)2.000)
= 0.848 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 25/4 = 6 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (6, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 25 = 400 mm
               iii) 300 mm
               Required spacing = minimum of (300, 400, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG14 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #25 - 14 nos. (6872 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG15 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 604.88 kN
MomentX,(Mx) = 13.89 kN-m
MomentY,(My) = 9.20 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C230.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 12.521 kN.m
My_MinEccen = 12.098 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 13.894 kN.m
My = max(My,My_MinEccen) + MuaddY = 12.098 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 13.89/604.88 = 23 mm
Actual eccenY = My / P = 9.20/604.88 = 15 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(23,21) = 23 mm
eccenY = max(Actual eccenY,eccenYMin) = max(15,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(23 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00287 352.79 8.92 343.87 77.78 179 13.92
2 226 0.00165 308.08 8.65 299.43 67.73 90 6.06
3 226 0.00043 86.30 3.43 82.87 18.74 0 0.00
4 226 -0.00079 -157.81 0.00 -157.81 -35.70 -90 3.19
5 226 -0.00201 -328.09 0.00 -328.09 -74.21 -179 13.28
Total54.34 36.46
xux = 257 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 257/450 = 0.205
Puc = 0.205 x 20.00 x 300 x 450
= 554.34 kN
Pux1 = Puc + Pus(Total)
= 554.34 + (54.34)
= 608.69 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 257/450
= 0.237
Muc = 554.34 x (0.5 x 450 - 0.237 x 450) = 65.54 kN-m
Mux1 = Muc + Mus(Total)
= 65.54 + (36.46)
= 102.01 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00260 347.74 8.92 338.82 191.60 104 19.93
2 565 -0.00146 -290.43 0.00 -290.43 -164.23 -104 17.08
Total27.36 37.01
xuy = 179 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 179/300 = 0.215
Puc = 0.215 x 20.00 x 450 x 300
= 580.92 kN
Puy1 = Puc + Pus(Total)
= 580.92 + (27.36)
= 608.29 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 179/300
= 0.249
Muc = 580.92 x (0.5 x 300 - 0.249 x 300) = 43.81 kN-m
Muy1 = Muc + Mus(Total)
= 43.81 + (37.01)
= 80.82 kN-m
Pu/Puz = 0.389
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.389, an = 1.314
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((13.89/102.01)1.314) + ((12.10/80.82)1.314)
= 0.155 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG15 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1214.02 kN
MomentX,(Mx) = 14.42 kN-m
MomentY,(My) = 8.59 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C230.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 25.130 kN.m
My_MinEccen = 24.280 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 25.130 kN.m
My = max(My,My_MinEccen) + MuaddY = 24.280 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 14.42/1214.02 = 12 mm
Actual eccenY = My / P = 8.59/1214.02 = 7 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(12,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(7,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00313 355.04 8.92 346.12 78.29 179 14.01
2 226 0.00241 342.85 8.92 333.93 75.53 90 6.76
3 226 0.00169 310.67 8.71 301.96 68.30 0 0.00
4 226 0.00097 195.00 6.58 188.42 42.62 -90 -3.81
5 226 0.00026 51.28 2.14 49.14 11.12 -179 -1.99
Total275.86 14.97
xux = 436 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 436/450 = 0.349
Puc = 0.349 x 20.00 x 300 x 450
= 941.63 kN
Pux1 = Puc + Pus(Total)
= 941.63 + (275.86)
= 1217.49 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 436/450
= 0.403
Muc = 941.63 x (0.5 x 450 - 0.403 x 450) = 41.10 kN-m
Mux1 = Muc + Mus(Total)
= 41.10 + (14.97)
= 56.07 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00296 353.58 8.92 344.66 194.90 104 20.27
2 565 0.00054 107.33 4.14 103.19 58.35 -104 -6.07
Total253.25 14.20
xuy = 300 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 300/300 = 0.360
Puc = 0.360 x 20.00 x 450 x 300
= 972.00 kN
Puy1 = Puc + Pus(Total)
= 972.00 + (253.25)
= 1225.25 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 300/300
= 0.416
Muc = 972.00 x (0.5 x 300 - 0.416 x 300) = 24.49 kN-m
Muy1 = Muc + Mus(Total)
= 24.49 + (14.20)
= 38.70 kN-m
Pu/Puz = 0.780
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.780, an = 1.966
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((25.13/56.07)1.966) + ((24.28/38.70)1.966)
= 0.606 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG15 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1803.60 kN
MomentX,(Mx) = 10.42 kN-m
MomentY,(My) = 7.70 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C230.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 37.334 kN.m
My_MinEccen = 36.072 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 37.334 kN.m
My = max(My,My_MinEccen) + MuaddY = 36.072 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 10.42/1803.60 = 6 mm
Actual eccenY = My / P = 7.70/1803.60 = 4 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(6,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(4,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 16 nos. (3217 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00313 355.01 8.92 346.09 139.17 177 24.63
2 402 0.00273 351.11 8.92 342.19 137.60 126 17.40
3 402 0.00234 340.23 8.92 331.31 133.23 76 10.11
4 402 0.00195 325.78 8.91 316.87 127.42 25 3.22
5 402 0.00155 299.44 8.48 290.96 117.00 -25 -2.96
6 402 0.00116 232.00 7.35 224.65 90.34 -76 -6.85
7 402 0.00077 153.33 5.53 147.81 59.44 -126 -7.51
8 402 0.00037 74.67 3.02 71.65 28.81 -177 -5.10
Total833.01 32.93
xux = 450 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 450/450 = 0.360
Puc = 0.360 x 20.00 x 300 x 450
= 972.00 kN
Pux1 = Puc + Pus(Total)
= 972.00 + (833.01)
= 1805.01 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 450/450
= 0.416
Muc = 972.00 x (0.5 x 450 - 0.416 x 450) = 36.74 kN-m
Mux1 = Muc + Mus(Total)
= 36.74 + (32.93)
= 69.67 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1608 0.00283 352.39 8.92 343.47 552.46 102 56.35
2 1608 0.00073 146.64 5.34 141.30 227.28 -102 -23.18
Total779.74 33.17
xuy = 323 mm               Puc = C1.fck.B.D
ku = 1.078
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.884
C1 = 0.446 x (1 - C3/6) = 0.380
Puc = 0.380 x 20.00 x 450 x 300
= 1026.69 kN
Puy1 = Puc + Pus(Total)
= 1026.69 + (779.74)
= 1806.43 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.438
Muc = 1026.69 x (0.5 x 300 - 0.438 x 300) = 19.02 kN-m
Muy1 = Muc + Mus(Total)
= 19.02 + (33.17)
= 52.19 kN-m
Pu/Puz = 0.825
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.825, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((37.33/69.67)2.000) + ((36.07/52.19)2.000)
= 0.765 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG15 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 16 nos. (3217 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 2364.75 kN
MomentX,(Mx) = 3.76 kN-m
MomentY,(My) = 6.67 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C230.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 48.950 kN.m
My_MinEccen = 47.295 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 48.950 kN.m
My = max(My,My_MinEccen) + MuaddY = 47.295 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 3.76/2364.75 = 2 mm
Actual eccenY = My / P = 6.67/2364.75 = 3 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(2,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(3,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #25 - 12 nos. (5890 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 982 0.00307 354.53 8.92 345.61 339.30 173 58.53
2 982 0.00251 345.33 8.92 336.41 330.27 104 34.18
3 982 0.00195 325.74 8.91 316.82 311.04 35 10.73
4 982 0.00138 276.48 8.07 268.41 263.51 -35 -9.09
5 982 0.00082 163.87 5.81 158.06 155.17 -104 -16.06
6 982 0.00026 51.26 2.14 49.12 48.22 -173 -8.32
Total1447.52 69.97
xux = 429 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 429/450 = 0.343
Puc = 0.343 x 20.00 x 300 x 450
= 926.44 kN
Pux1 = Puc + Pus(Total)
= 926.44 + (1447.52)
= 2373.95 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 429/450
= 0.397
Muc = 926.44 x (0.5 x 450 - 0.397 x 450) = 43.15 kN-m
Mux1 = Muc + Mus(Total)
= 43.15 + (69.97)
= 113.12 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 2945 0.00286 352.71 8.92 343.79 1012.54 98 98.72
2 2945 0.00065 129.89 4.85 125.03 368.25 -98 -35.90
Total1380.80 62.82
xuy = 305 mm               Puc = C1.fck.B.D
ku = 1.016
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.083
C1 = 0.446 x (1 - C3/6) = 0.366
Puc = 0.366 x 20.00 x 450 x 300
= 986.88 kN
Puy1 = Puc + Pus(Total)
= 986.88 + (1380.80)
= 2367.67 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.421
Muc = 986.88 x (0.5 x 300 - 0.421 x 300) = 23.28 kN-m
Muy1 = Muc + Mus(Total)
= 23.28 + (62.82)
= 86.10 kN-m
Pu/Puz = 0.789
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.789, an = 1.982
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((48.95/113.12)1.982) + ((47.30/86.10)1.982)
= 0.495 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 25/4 = 6 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (6, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 25 = 400 mm
               iii) 300 mm
               Required spacing = minimum of (300, 400, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG15 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #25 - 12 nos. (5890 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 2859.01 kN
MomentX,(Mx) = 0.84 kN-m
MomentY,(My) = 4.04 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C230.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 62.184 kN.m
My_MinEccen = 57.180 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 62.184 kN.m
My = max(My,My_MinEccen) + MuaddY = 57.180 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 0.84/2859.01 = 0 mm
Actual eccenY = My / P = 4.04/2859.01 = 1 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(0,22) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(1,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #32 - 10 nos. (8042 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1608 0.00304 354.24 8.92 345.32 555.45 169 93.87
2 1608 0.00234 340.38 8.92 331.46 533.14 85 45.05
3 1608 0.00165 307.87 8.64 299.23 481.30 0 0.00
4 1608 0.00095 190.70 6.48 184.23 296.33 -85 -25.04
5 1608 0.00026 51.65 2.16 49.50 79.62 -169 -13.46
Total1945.85 100.43
xux = 425 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 425/450 = 0.340
Puc = 0.340 x 20.00 x 300 x 450
= 918.84 kN
Pux1 = Puc + Pus(Total)
= 918.84 + (1945.85)
= 2864.69 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 425/450
= 0.393
Muc = 918.84 x (0.5 x 450 - 0.393 x 450) = 44.14 kN-m
Mux1 = Muc + Mus(Total)
= 44.14 + (100.43)
= 144.57 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 4021 0.00285 352.56 8.92 343.64 1381.85 94 129.89
2 4021 0.00065 130.67 4.88 125.79 505.83 -94 -47.55
Total1887.69 82.35
xuy = 300 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 300/300 = 0.360
Puc = 0.360 x 20.00 x 450 x 300
= 972.00 kN
Puy1 = Puc + Pus(Total)
= 972.00 + (1887.69)
= 2859.69 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 300/300
= 0.416
Muc = 972.00 x (0.5 x 300 - 0.416 x 300) = 24.49 kN-m
Muy1 = Muc + Mus(Total)
= 24.49 + (82.35)
= 106.84 kN-m
Pu/Puz = 0.784
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.784, an = 1.974
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((62.18/144.57)1.974) + ((57.18/106.84)1.974)
= 0.480 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 32/4 = 8 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (8, 6) = 8 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 32 = 512 mm
               iii) 300 mm
               Required spacing = minimum of (300, 512, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG15 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #32 - 10 nos. (8042 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG16 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 650.80 kN
MomentX,(Mx) = 62.68 kN-m
MomentY,(My) = 2.05 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C240.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 13.472 kN.m
My_MinEccen = 13.016 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 62.678 kN.m
My = max(My,My_MinEccen) + MuaddY = 13.016 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 62.68/650.80 = 96 mm
Actual eccenY = My / P = 2.05/650.80 = 3 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(96,21) = 96 mm
eccenY = max(Actual eccenY,eccenYMin) = max(3,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(96 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00290 353.04 8.92 344.12 77.84 179 13.93
2 226 0.00174 313.36 8.77 304.59 68.90 90 6.17
3 226 0.00057 114.38 4.37 110.01 24.88 0 0.00
4 226 -0.00059 -118.57 0.00 -118.57 -26.82 -90 2.40
5 226 -0.00176 -314.66 0.00 -314.66 -71.17 -179 12.74
Total73.62 35.24
xux = 269 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 269/450 = 0.215
Puc = 0.215 x 20.00 x 300 x 450
= 580.92 kN
Pux1 = Puc + Pus(Total)
= 580.92 + (73.62)
= 654.54 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 269/450
= 0.249
Muc = 580.92 x (0.5 x 450 - 0.249 x 450) = 65.71 kN-m
Mux1 = Muc + Mus(Total)
= 65.71 + (35.24)
= 100.95 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00264 348.61 8.92 339.69 192.09 104 19.98
2 565 -0.00127 -254.23 0.00 -254.23 -143.76 -104 14.95
Total48.33 34.93
xuy = 186 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 186/300 = 0.224
Puc = 0.224 x 20.00 x 450 x 300
= 603.70 kN
Puy1 = Puc + Pus(Total)
= 603.70 + (48.33)
= 652.03 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 186/300
= 0.258
Muc = 603.70 x (0.5 x 300 - 0.258 x 300) = 43.76 kN-m
Muy1 = Muc + Mus(Total)
= 43.76 + (34.93)
= 78.69 kN-m
Pu/Puz = 0.418
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.418, an = 1.363
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((62.68/100.95)1.363) + ((13.02/78.69)1.363)
= 0.608 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG16 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1301.16 kN
MomentX,(Mx) = 54.20 kN-m
MomentY,(My) = 1.06 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C240.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 26.934 kN.m
My_MinEccen = 26.023 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 54.200 kN.m
My = max(My,My_MinEccen) + MuaddY = 26.023 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 54.20/1301.16 = 42 mm
Actual eccenY = My / P = 1.06/1301.16 = 1 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(42,21) = 42 mm
eccenY = max(Actual eccenY,eccenYMin) = max(1,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(42 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 8 nos. (1608 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00311 354.85 8.92 345.93 139.11 177 24.62
2 402 0.00215 333.08 8.92 324.16 130.35 59 7.69
3 402 0.00118 236.50 7.43 229.07 92.11 -59 -5.43
4 402 0.00022 43.91 1.85 42.06 16.91 -177 -2.99
Total378.48 23.88
xux = 429 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 429/450 = 0.343
Puc = 0.343 x 20.00 x 300 x 450
= 926.44 kN
Pux1 = Puc + Pus(Total)
= 926.44 + (378.48)
= 1304.92 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 429/450
= 0.397
Muc = 926.44 x (0.5 x 450 - 0.397 x 450) = 43.15 kN-m
Mux1 = Muc + Mus(Total)
= 43.15 + (23.88)
= 67.03 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 804 0.00293 353.30 8.92 344.38 276.96 102 28.25
2 804 0.00051 102.67 3.99 98.68 79.36 -102 -8.09
Total356.32 20.16
xuy = 295 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 295/300 = 0.354
Puc = 0.354 x 20.00 x 450 x 300
= 956.81 kN
Puy1 = Puc + Pus(Total)
= 956.81 + (356.32)
= 1313.14 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 295/300
= 0.410
Muc = 956.81 x (0.5 x 300 - 0.410 x 300) = 25.98 kN-m
Muy1 = Muc + Mus(Total)
= 25.98 + (20.16)
= 46.13 kN-m
Pu/Puz = 0.765
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.765, an = 1.941
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((54.20/67.03)1.941) + ((26.02/46.13)1.941)
= 0.991 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG16 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 8 nos. (1608 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1973.48 kN
MomentX,(Mx) = 42.96 kN-m
MomentY,(My) = 1.21 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C240.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 40.851 kN.m
My_MinEccen = 39.470 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 42.961 kN.m
My = max(My,My_MinEccen) + MuaddY = 39.470 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 42.96/1973.48 = 22 mm
Actual eccenY = My / P = 1.21/1973.48 = 1 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(22,21) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(1,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 12 nos. (3770 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00305 354.37 8.92 345.45 217.05 175 37.98
2 628 0.00254 346.07 8.92 337.15 211.84 105 22.24
3 628 0.00202 328.51 8.92 319.59 200.81 35 7.03
4 628 0.00150 294.84 8.37 286.47 180.00 -35 -6.30
5 628 0.00099 197.73 6.64 191.09 120.06 -105 -12.61
6 628 0.00047 94.49 3.72 90.77 57.03 -175 -9.98
Total986.79 38.37
xux = 464 mm               Puc = C1.fck.B.D
ku = 1.031
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.027
C1 = 0.446 x (1 - C3/6) = 0.370
Puc = 0.370 x 20.00 x 300 x 450
= 998.00 kN
Pux1 = Puc + Pus(Total)
= 998.00 + (986.79)
= 1984.79 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.426
Muc = 998.00 x (0.5 x 450 - 0.426 x 450) = 33.14 kN-m
Mux1 = Muc + Mus(Total)
= 33.14 + (38.37)
= 71.51 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1885 0.00277 351.88 8.92 342.96 646.47 100 64.65
2 1885 0.00081 162.19 5.77 156.42 294.84 -100 -29.48
Total941.31 35.16
xuy = 333 mm               Puc = C1.fck.B.D
ku = 1.109
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.805
C1 = 0.446 x (1 - C3/6) = 0.386
Puc = 0.386 x 20.00 x 450 x 300
= 1042.61 kN
Puy1 = Puc + Pus(Total)
= 1042.61 + (941.31)
= 1983.92 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.445
Muc = 1042.61 x (0.5 x 300 - 0.445 x 300) = 17.31 kN-m
Muy1 = Muc + Mus(Total)
= 17.31 + (35.16)
= 52.48 kN-m
Pu/Puz = 0.838
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.838, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((42.96/71.51)2.000) + ((39.47/52.48)2.000)
= 0.927 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 20 = 320 mm
               iii) 300 mm
               Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG16 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 12 nos. (3770 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 2598.35 kN
MomentX,(Mx) = 59.23 kN-m
MomentY,(My) = 4.06 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C240.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 53.786 kN.m
My_MinEccen = 51.967 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 59.228 kN.m
My = max(My,My_MinEccen) + MuaddY = 51.967 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 59.23/2598.35 = 23 mm
Actual eccenY = My / P = 4.06/2598.35 = 2 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(23,21) = 23 mm
eccenY = max(Actual eccenY,eccenYMin) = max(2,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(23 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #25 - 14 nos. (6872 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 982 0.00307 354.53 8.92 345.61 339.30 173 58.53
2 982 0.00260 347.75 8.92 338.83 332.64 115 38.25
3 982 0.00213 332.63 8.92 323.71 317.80 58 18.27
4 982 0.00166 308.82 8.67 300.15 294.67 0 0.00
5 982 0.00119 238.94 7.47 231.47 227.24 -58 -13.07
6 982 0.00073 145.10 5.30 139.80 137.25 -115 -15.78
7 982 0.00026 51.26 2.14 49.12 48.22 -173 -8.32
Total1697.13 77.89
xux = 429 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 429/450 = 0.343
Puc = 0.343 x 20.00 x 300 x 450
= 926.44 kN
Pux1 = Puc + Pus(Total)
= 926.44 + (1697.13)
= 2623.57 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 429/450
= 0.397
Muc = 926.44 x (0.5 x 450 - 0.397 x 450) = 43.15 kN-m
Mux1 = Muc + Mus(Total)
= 43.15 + (77.89)
= 121.04 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 3436 0.00285 352.61 8.92 343.69 1180.96 98 115.14
2 3436 0.00067 133.43 4.96 128.47 441.45 -98 -43.04
Total1622.41 72.10
xuy = 307 mm               Puc = C1.fck.B.D
ku = 1.023
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.055
C1 = 0.446 x (1 - C3/6) = 0.368
Puc = 0.368 x 20.00 x 450 x 300
= 992.55 kN
Puy1 = Puc + Pus(Total)
= 992.55 + (1622.41)
= 2614.96 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.424
Muc = 992.55 x (0.5 x 300 - 0.424 x 300) = 22.68 kN-m
Muy1 = Muc + Mus(Total)
= 22.68 + (72.10)
= 94.78 kN-m
Pu/Puz = 0.789
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.789, an = 1.982
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((59.23/121.04)1.982) + ((51.97/94.78)1.982)
= 0.546 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 25/4 = 6 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (6, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 25 = 400 mm
               iii) 300 mm
               Required spacing = minimum of (300, 400, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG16 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #25 - 14 nos. (6872 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 3348.72 kN
MomentX,(Mx) = 50.85 kN-m
MomentY,(My) = 3.56 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C240.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 72.835 kN.m
My_MinEccen = 66.974 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 72.835 kN.m
My = max(My,My_MinEccen) + MuaddY = 66.974 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 50.85/3348.72 = 15 mm
Actual eccenY = My / P = 3.56/3348.72 = 1 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(15,22) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(1,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #32 - 12 nos. (9651 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1608 0.00305 354.34 8.92 345.42 555.61 169 93.90
2 1608 0.00251 345.31 8.92 336.39 541.08 101 54.87
3 1608 0.00196 326.45 8.92 317.53 510.75 34 17.26
4 1608 0.00142 284.44 8.18 276.26 444.36 -34 -15.02
5 1608 0.00088 175.89 6.12 169.77 273.07 -101 -27.69
6 1608 0.00034 67.34 2.75 64.59 103.89 -169 -17.56
Total2428.77 105.76
xux = 436 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 436/450 = 0.349
Puc = 0.349 x 20.00 x 300 x 450
= 941.63 kN
Pux1 = Puc + Pus(Total)
= 941.63 + (2428.77)
= 3370.39 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 436/450
= 0.403
Muc = 941.63 x (0.5 x 450 - 0.403 x 450) = 41.10 kN-m
Mux1 = Muc + Mus(Total)
= 41.10 + (105.76)
= 146.86 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 4825 0.00278 352.00 8.92 343.08 1655.51 94 155.62
2 4825 0.00076 151.09 5.47 145.62 702.69 -94 -66.05
Total2358.20 89.57
xuy = 314 mm               Puc = C1.fck.B.D
ku = 1.047
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.976
C1 = 0.446 x (1 - C3/6) = 0.373
Puc = 0.373 x 20.00 x 450 x 300
= 1008.29 kN
Puy1 = Puc + Pus(Total)
= 1008.29 + (2358.20)
= 3366.49 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.431
Muc = 1008.29 x (0.5 x 300 - 0.431 x 300) = 20.99 kN-m
Muy1 = Muc + Mus(Total)
= 20.99 + (89.57)
= 110.56 kN-m
Pu/Puz = 0.810
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.810, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((72.83/146.86)2.000) + ((66.97/110.56)2.000)
= 0.613 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 32/4 = 8 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (8, 6) = 8 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 32 = 512 mm
               iii) 300 mm
               Required spacing = minimum of (300, 512, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG16 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #32 - 12 nos. (9651 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG17 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3450 mm
From analysis results,loads on column
Axial load,(P) = 469.48 kN
MomentX,(Mx) = 62.65 kN-m
MomentY,(My) = 31.76 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C250.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3450 x 1.000 ) / 450 = 7.67 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3450 x 1.000 ) / 300 = 11.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 10.282 kN.m
My_MinEccen = 9.390 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 62.649 kN.m
My = max(My,My_MinEccen) + MuaddY = 31.757 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 62.65/469.48 = 133 mm
Actual eccenY = My / P = 31.76/469.48 = 68 mm
eccenXMin = (L/500) + (D/30) = (3450/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3450/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(133,22) = 133 mm
eccenY = max(Actual eccenY,eccenYMin) = max(68,20) = 68 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(133 mm) > 0.05 x 300(15 mm)
and eccenY(68 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00278 351.94 8.92 343.02 77.59 179 13.89
2 226 0.00137 273.45 8.03 265.42 60.04 90 5.37
3 226 -0.00004 -8.30 0.00 -8.30 -1.88 0 0.00
4 226 -0.00145 -289.67 0.00 -289.67 -65.52 -90 5.86
5 226 -0.00286 -352.67 0.00 -352.67 -79.77 -179 14.28
Total-9.54 39.40
xux = 222 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 222/450 = 0.178
Puc = 0.178 x 20.00 x 300 x 450
= 480.30 kN
Pux1 = Puc + Pus(Total)
= 480.30 + (-9.54)
= 470.76 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 222/450
= 0.206
Muc = 480.30 x (0.5 x 450 - 0.206 x 450) = 63.64 kN-m
Mux1 = Muc + Mus(Total)
= 63.64 + (39.40)
= 103.04 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00241 342.79 8.92 333.87 188.80 104 19.63
2 565 -0.00252 -345.65 0.00 -345.65 -195.46 -104 20.33
Total-6.66 39.96
xuy = 148 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 148/300 = 0.177
Puc = 0.177 x 20.00 x 450 x 300
= 478.41 kN
Puy1 = Puc + Pus(Total)
= 478.41 + (-6.66)
= 471.74 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 148/300
= 0.205
Muc = 478.41 x (0.5 x 300 - 0.205 x 300) = 42.37 kN-m
Muy1 = Muc + Mus(Total)
= 42.37 + (39.96)
= 82.34 kN-m
Pu/Puz = 0.302
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.302, an = 1.169
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((62.65/103.04)1.169) + ((31.76/82.34)1.169)
= 0.887 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG17 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3450 mm
From analysis results,loads on column
Axial load,(P) = 936.89 kN
MomentX,(Mx) = 38.97 kN-m
MomentY,(My) = 29.31 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C250.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3450 x 1.000 ) / 450 = 7.67 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3450 x 1.000 ) / 300 = 11.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 20.518 kN.m
My_MinEccen = 18.738 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 38.969 kN.m
My = max(My,My_MinEccen) + MuaddY = 29.313 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 38.97/936.89 = 42 mm
Actual eccenY = My / P = 29.31/936.89 = 31 mm
eccenXMin = (L/500) + (D/30) = (3450/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3450/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(42,22) = 42 mm
eccenY = max(Actual eccenY,eccenYMin) = max(31,20) = 31 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(42 mm) > 0.05 x 300(15 mm)
and eccenY(31 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00304 354.21 8.92 345.29 78.10 179 13.98
2 226 0.00213 332.53 8.92 323.61 73.20 90 6.55
3 226 0.00123 245.18 7.58 237.59 53.74 0 0.00
4 226 0.00032 64.26 2.64 61.62 13.94 -90 -1.25
5 226 -0.00058 -116.66 0.00 -116.66 -26.39 -179 4.72
Total192.60 24.01
xux = 346 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 346/450 = 0.277
Puc = 0.277 x 20.00 x 300 x 450
= 747.98 kN
Pux1 = Puc + Pus(Total)
= 747.98 + (192.60)
= 940.58 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 346/450
= 0.320
Muc = 747.98 x (0.5 x 450 - 0.320 x 450) = 60.54 kN-m
Mux1 = Muc + Mus(Total)
= 60.54 + (24.01)
= 84.55 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00283 352.38 8.92 343.46 194.22 104 20.20
2 565 -0.00022 -43.74 0.00 -43.74 -24.73 -104 2.57
Total169.49 22.77
xuy = 239 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 239/300 = 0.287
Puc = 0.287 x 20.00 x 450 x 300
= 774.56 kN
Puy1 = Puc + Pus(Total)
= 774.56 + (169.49)
= 944.05 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 239/300
= 0.331
Muc = 774.56 x (0.5 x 300 - 0.331 x 300) = 39.15 kN-m
Muy1 = Muc + Mus(Total)
= 39.15 + (22.77)
= 61.93 kN-m
Pu/Puz = 0.602
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.602, an = 1.670
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((38.97/84.55)1.670) + ((29.31/61.93)1.670)
= 0.561 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG17 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3150 mm
From analysis results,loads on column
Axial load,(P) = 1394.40 kN
MomentX,(Mx) = 39.09 kN-m
MomentY,(My) = 25.97 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C250.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3150 x 1.000 ) / 450 = 7.00 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3150 x 1.000 ) / 300 = 10.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 29.701 kN.m
My_MinEccen = 27.888 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 39.095 kN.m
My = max(My,My_MinEccen) + MuaddY = 27.888 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 39.09/1394.40 = 28 mm
Actual eccenY = My / P = 25.97/1394.40 = 19 mm
eccenXMin = (L/500) + (D/30) = (3150/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (3150/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(28,21) = 28 mm
eccenY = max(Actual eccenY,eccenYMin) = max(19,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(28 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 4 nos. + #16 - 6 nos. (1659 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00310 354.75 8.92 345.83 139.06 177 24.61
2 226 0.00243 343.23 8.92 334.31 75.62 89 6.69
3 402 0.00176 314.60 8.79 305.82 122.98 0 0.00
4 226 0.00109 217.33 7.06 210.27 47.56 -89 -4.21
5 402 0.00042 83.33 3.33 80.00 32.17 -177 -5.69
Total417.39 21.40
xux = 457 mm               Puc = C1.fck.B.D
ku = 1.016
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.083
C1 = 0.446 x (1 - C3/6) = 0.366
Puc = 0.366 x 20.00 x 300 x 450
= 986.88 kN
Pux1 = Puc + Pus(Total)
= 986.88 + (417.39)
= 1404.27 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.421
Muc = 986.88 x (0.5 x 450 - 0.421 x 450) = 34.93 kN-m
Mux1 = Muc + Mus(Total)
= 34.93 + (21.40)
= 56.33 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 829 0.00287 352.75 8.92 343.83 285.17 102 29.09
2 829 0.00067 133.83 4.97 128.86 106.88 -102 -10.90
Total392.04 18.19
xuy = 314 mm               Puc = C1.fck.B.D
ku = 1.047
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.976
C1 = 0.446 x (1 - C3/6) = 0.373
Puc = 0.373 x 20.00 x 450 x 300
= 1008.29 kN
Puy1 = Puc + Pus(Total)
= 1008.29 + (392.04)
= 1400.33 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.431
Muc = 1008.29 x (0.5 x 300 - 0.431 x 300) = 20.99 kN-m
Muy1 = Muc + Mus(Total)
= 20.99 + (18.19)
= 39.18 kN-m
Pu/Puz = 0.812
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.812, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((39.09/56.33)2.000) + ((27.89/39.18)2.000)
= 0.988 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG17 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 4 nos. + #16 - 6 nos. (1659 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3150 mm
From analysis results,loads on column
Axial load,(P) = 1903.29 kN
MomentX,(Mx) = 45.03 kN-m
MomentY,(My) = 20.53 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C250.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3150 x 1.000 ) / 450 = 7.00 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3150 x 1.000 ) / 300 = 10.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 40.540 kN.m
My_MinEccen = 38.066 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 45.031 kN.m
My = max(My,My_MinEccen) + MuaddY = 38.066 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 45.03/1903.29 = 24 mm
Actual eccenY = My / P = 20.53/1903.29 = 11 mm
eccenXMin = (L/500) + (D/30) = (3150/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (3150/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(24,21) = 24 mm
eccenY = max(Actual eccenY,eccenYMin) = max(11,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(24 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 18 nos. (3619 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00313 355.01 8.92 346.09 139.17 177 24.63
2 402 0.00278 352.00 8.92 343.08 137.96 133 18.31
3 402 0.00244 343.53 8.92 334.61 134.55 89 11.91
4 402 0.00209 331.20 8.92 322.28 129.60 44 5.73
5 402 0.00175 314.19 8.78 305.41 122.81 0 0.00
6 402 0.00141 281.17 8.13 273.03 109.79 -44 -4.86
7 402 0.00106 212.33 6.96 205.38 82.59 -89 -7.31
8 402 0.00072 143.50 5.25 138.25 55.59 -133 -7.38
9 402 0.00037 74.67 3.02 71.65 28.81 -177 -5.10
Total940.88 35.94
xux = 450 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 450/450 = 0.360
Puc = 0.360 x 20.00 x 300 x 450
= 972.00 kN
Pux1 = Puc + Pus(Total)
= 972.00 + (940.88)
= 1912.88 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 450/450
= 0.416
Muc = 972.00 x (0.5 x 450 - 0.416 x 450) = 36.74 kN-m
Mux1 = Muc + Mus(Total)
= 36.74 + (35.94)
= 72.68 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1810 0.00281 352.22 8.92 343.30 621.21 102 63.36
2 1810 0.00076 152.59 5.51 147.08 266.16 -102 -27.15
Total887.37 36.22
xuy = 328 mm               Puc = C1.fck.B.D
ku = 1.094
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.843
C1 = 0.446 x (1 - C3/6) = 0.383
Puc = 0.383 x 20.00 x 450 x 300
= 1034.93 kN
Puy1 = Puc + Pus(Total)
= 1034.93 + (887.37)
= 1922.30 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.442
Muc = 1034.93 x (0.5 x 300 - 0.442 x 300) = 18.14 kN-m
Muy1 = Muc + Mus(Total)
= 18.14 + (36.22)
= 54.35 kN-m
Pu/Puz = 0.824
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.824, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((45.03/72.68)2.000) + ((38.07/54.35)2.000)
= 0.874 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG17 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 18 nos. (3619 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3675 mm
From analysis results,loads on column
Axial load,(P) = 2366.40 kN
MomentX,(Mx) = 31.83 kN-m
MomentY,(My) = 10.87 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C250.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3675 x 1.000 ) / 450 = 8.17 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3675 x 1.000 ) / 300 = 12.25 > 12.00,
column is slender in this direction.
Puz = (0.45 x fck x Ag) + (0.75 x fy - 0.45 x fck) x Ast
= 0.45 x 20.00 x 135000.01 + (0.75 x 415.00 - 0.45 x 20.00 ) x 5890.49 = 2995399.658 N
k1 = 0.190000
k2 = 0.012000 MPa
Pub = (k1 x fck x Ag) + (k2 x Ast x 100)
= (0.190000 x 20.00 x 135000.01) + (0.01 x 5890.49 x 100) = 520068.604 N
k = (Puz - Pu) / (Puz - Pub) <= 1
= (2995399.658 - 2366396.973) / (2995399.658 - 520068.604) = 0.254109
MuaddY = Pu x (Width/2000) x (Lo/Width) x (Lo/Width) x k
= 2366396.97 x (300.00/2000) x (3675.00/300.00) x (3675.00/300.00) x 0.254109
= 13535374.641 N.mm = 13.535 kN.m
Mx_MinEccen = 52.889 kN.m
My_MinEccen = 47.328 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 52.889 kN.m
My = max(My,My_MinEccen) + MuaddY = 60.863 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 31.83/2366.40 = 13 mm
Actual eccenY = My / P = 10.87/2366.40 = 5 mm
eccenXMin = (L/500) + (D/30) = (3675/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3675/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(13,22) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(5,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #25 - 12 nos. (5890 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 982 0.00307 354.53 8.92 345.61 339.30 173 58.53
2 982 0.00251 345.33 8.92 336.41 330.27 104 34.18
3 982 0.00195 325.74 8.91 316.82 311.04 35 10.73
4 982 0.00138 276.48 8.07 268.41 263.51 -35 -9.09
5 982 0.00082 163.87 5.81 158.06 155.17 -104 -16.06
6 982 0.00026 51.26 2.14 49.12 48.22 -173 -8.32
Total1447.52 69.97
xux = 429 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 429/450 = 0.343
Puc = 0.343 x 20.00 x 300 x 450
= 926.44 kN
Pux1 = Puc + Pus(Total)
= 926.44 + (1447.52)
= 2373.95 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 429/450
= 0.397
Muc = 926.44 x (0.5 x 450 - 0.397 x 450) = 43.15 kN-m
Mux1 = Muc + Mus(Total)
= 43.15 + (69.97)
= 113.12 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 2945 0.00286 352.71 8.92 343.79 1012.54 98 98.72
2 2945 0.00065 129.89 4.85 125.03 368.25 -98 -35.90
Total1380.80 62.82
xuy = 305 mm               Puc = C1.fck.B.D
ku = 1.016
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.083
C1 = 0.446 x (1 - C3/6) = 0.366
Puc = 0.366 x 20.00 x 450 x 300
= 986.88 kN
Puy1 = Puc + Pus(Total)
= 986.88 + (1380.80)
= 2367.67 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.421
Muc = 986.88 x (0.5 x 300 - 0.421 x 300) = 23.28 kN-m
Muy1 = Muc + Mus(Total)
= 23.28 + (62.82)
= 86.10 kN-m
Pu/Puz = 0.790
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.790, an = 1.983
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((52.89/113.12)1.983) + ((60.86/86.10)1.983)
= 0.724 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 25/4 = 6 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (6, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 25 = 400 mm
               iii) 300 mm
               Required spacing = minimum of (300, 400, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG17 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #25 - 12 nos. (5890 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG18 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 414.57 kN
MomentX,(Mx) = 96.63 kN-m
MomentY,(My) = 2.62 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C260.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 8.581 kN.m
My_MinEccen = 8.291 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 96.627 kN.m
My = max(My,My_MinEccen) + MuaddY = 8.291 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 96.63/414.57 = 233 mm
Actual eccenY = My / P = 2.62/414.57 = 6 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(233,21) = 233 mm
eccenY = max(Actual eccenY,eccenYMin) = max(6,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(233 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 6 nos. (1206 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00269 350.00 8.92 341.08 137.16 177 24.28
2 402 -0.00030 -59.32 0.00 -59.32 -23.85 0 0.00
3 402 -0.00328 -356.38 0.00 -356.38 -143.31 -177 25.37
Total-30.01 49.64
xux = 207 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 207/450 = 0.166
Puc = 0.166 x 20.00 x 300 x 450
= 448.03 kN
Pux1 = Puc + Pus(Total)
= 448.03 + (-30.01)
= 418.03 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 207/450
= 0.192
Muc = 448.03 x (0.5 x 450 - 0.192 x 450) = 62.15 kN-m
Mux1 = Muc + Mus(Total)
= 62.15 + (49.64)
= 111.79 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 603 0.00224 336.65 8.92 327.73 197.68 102 20.16
2 603 -0.00310 -354.79 0.00 -354.79 -214.01 -102 21.83
Total-16.33 41.99
xuy = 134 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 134/300 = 0.160
Puc = 0.160 x 20.00 x 450 x 300
= 432.84 kN
Puy1 = Puc + Pus(Total)
= 432.84 + (-16.33)
= 416.52 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 134/300
= 0.185
Muc = 432.84 x (0.5 x 300 - 0.185 x 300) = 40.87 kN-m
Muy1 = Muc + Mus(Total)
= 40.87 + (41.99)
= 82.86 kN-m
Pu/Puz = 0.262
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.262, an = 1.104
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((96.63/111.79)1.104) + ((8.29/82.86)1.104)
= 0.930 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG18 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 6 nos. (1206 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 877.60 kN
MomentX,(Mx) = 89.66 kN-m
MomentY,(My) = 2.43 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C260.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 18.166 kN.m
My_MinEccen = 17.552 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 89.660 kN.m
My = max(My,My_MinEccen) + MuaddY = 17.552 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 89.66/877.60 = 102 mm
Actual eccenY = My / P = 2.43/877.60 = 3 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(102,21) = 102 mm
eccenY = max(Actual eccenY,eccenYMin) = max(3,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(102 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 6 nos. (1206 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00299 353.83 8.92 344.91 138.70 177 24.55
2 402 0.00112 223.40 7.18 216.22 86.95 0 0.00
3 402 -0.00076 -151.52 0.00 -151.52 -60.93 -177 10.78
Total164.71 35.33
xux = 330 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 330/450 = 0.264
Puc = 0.264 x 20.00 x 300 x 450
= 713.81 kN
Pux1 = Puc + Pus(Total)
= 713.81 + (164.71)
= 878.53 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 330/450
= 0.306
Muc = 713.81 x (0.5 x 450 - 0.306 x 450) = 62.48 kN-m
Mux1 = Muc + Mus(Total)
= 62.48 + (35.33)
= 97.81 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 603 0.00275 351.63 8.92 342.71 206.72 102 21.09
2 603 -0.00042 -84.00 0.00 -84.00 -50.67 -102 5.17
Total156.05 26.25
xuy = 225 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 225/300 = 0.270
Puc = 0.270 x 20.00 x 450 x 300
= 729.00 kN
Puy1 = Puc + Pus(Total)
= 729.00 + (156.05)
= 885.05 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 225/300
= 0.312
Muc = 729.00 x (0.5 x 300 - 0.312 x 300) = 41.12 kN-m
Muy1 = Muc + Mus(Total)
= 41.12 + (26.25)
= 67.37 kN-m
Pu/Puz = 0.556
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.556, an = 1.593
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((89.66/97.81)1.593) + ((17.55/67.37)1.593)
= 0.988 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG18 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 6 nos. (1206 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1307.05 kN
MomentX,(Mx) = 83.60 kN-m
MomentY,(My) = 2.38 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C260.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 27.056 kN.m
My_MinEccen = 26.141 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 83.596 kN.m
My = max(My,My_MinEccen) + MuaddY = 26.141 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 83.60/1307.05 = 64 mm
Actual eccenY = My / P = 2.38/1307.05 = 2 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(64,21) = 64 mm
eccenY = max(Actual eccenY,eccenYMin) = max(2,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(64 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 12 nos. (2413 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00306 354.40 8.92 345.48 138.93 177 24.59
2 402 0.00240 342.61 8.92 333.69 134.19 106 14.25
3 402 0.00175 314.33 8.78 305.55 122.87 35 4.35
4 402 0.00110 219.92 7.11 212.81 85.58 -35 -3.03
5 402 0.00045 89.39 3.54 85.85 34.52 -106 -3.67
6 402 -0.00021 -41.14 0.00 -41.14 -16.54 -177 2.93
Total499.54 39.42
xux = 380 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 380/450 = 0.304
Puc = 0.304 x 20.00 x 300 x 450
= 820.13 kN
Pux1 = Puc + Pus(Total)
= 820.13 + (499.54)
= 1319.66 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 380/450
= 0.351
Muc = 820.13 x (0.5 x 450 - 0.351 x 450) = 54.99 kN-m
Mux1 = Muc + Mus(Total)
= 54.99 + (39.42)
= 94.41 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1206 0.00287 352.72 8.92 343.80 414.76 102 42.31
2 1206 0.00017 33.95 1.45 32.50 39.20 -102 -4.00
Total453.96 38.31
xuy = 265 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 265/300 = 0.318
Puc = 0.318 x 20.00 x 450 x 300
= 858.09 kN
Puy1 = Puc + Pus(Total)
= 858.09 + (453.96)
= 1312.05 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 265/300
= 0.367
Muc = 858.09 x (0.5 x 300 - 0.367 x 300) = 34.17 kN-m
Muy1 = Muc + Mus(Total)
= 34.17 + (38.31)
= 72.48 kN-m
Pu/Puz = 0.672
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.672, an = 1.787
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((83.60/94.41)1.787) + ((26.14/72.48)1.787)
= 0.966 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG18 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 12 nos. (2413 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1694.50 kN
MomentX,(Mx) = 71.75 kN-m
MomentY,(My) = 2.12 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C260.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 35.076 kN.m
My_MinEccen = 33.890 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 71.752 kN.m
My = max(My,My_MinEccen) + MuaddY = 33.890 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 71.75/1694.50 = 42 mm
Actual eccenY = My / P = 2.12/1694.50 = 1 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(42,21) = 42 mm
eccenY = max(Actual eccenY,eccenYMin) = max(1,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(42 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 18 nos. (3619 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00308 354.61 8.92 345.69 139.01 177 24.60
2 402 0.00269 350.11 8.92 341.19 137.20 133 18.21
3 402 0.00231 339.05 8.92 330.13 132.75 89 11.75
4 402 0.00192 324.86 8.91 315.95 127.05 44 5.62
5 402 0.00154 297.71 8.44 289.27 116.32 0 0.00
6 402 0.00115 229.73 7.30 222.43 89.44 -44 -3.96
7 402 0.00076 152.44 5.50 146.94 59.09 -89 -5.23
8 402 0.00038 75.16 3.04 72.12 29.00 -133 -3.85
9 402 -0.00001 -2.13 0.00 -2.13 -0.86 -177 0.15
Total829.02 47.30
xux = 401 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 401/450 = 0.321
Puc = 0.321 x 20.00 x 300 x 450
= 865.69 kN
Pux1 = Puc + Pus(Total)
= 865.69 + (829.02)
= 1694.70 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 401/450
= 0.370
Muc = 865.69 x (0.5 x 450 - 0.370 x 450) = 50.45 kN-m
Mux1 = Muc + Mus(Total)
= 50.45 + (47.30)
= 97.75 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1810 0.00292 353.18 8.92 344.26 622.95 102 63.54
2 1810 0.00044 88.10 3.50 84.60 153.09 -102 -15.62
Total776.04 47.93
xuy = 288 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 288/300 = 0.346
Puc = 0.346 x 20.00 x 450 x 300
= 934.03 kN
Puy1 = Puc + Pus(Total)
= 934.03 + (776.04)
= 1710.07 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 288/300
= 0.400
Muc = 934.03 x (0.5 x 300 - 0.400 x 300) = 28.09 kN-m
Muy1 = Muc + Mus(Total)
= 28.09 + (47.93)
= 76.02 kN-m
Pu/Puz = 0.734
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.734, an = 1.890
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((71.75/97.75)1.890) + ((33.89/76.02)1.890)
= 0.775 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG18 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 18 nos. (3619 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 1904.43 kN
MomentX,(Mx) = 43.14 kN-m
MomentY,(My) = 0.72 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C260.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 41.421 kN.m
My_MinEccen = 38.089 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 43.136 kN.m
My = max(My,My_MinEccen) + MuaddY = 38.089 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 43.14/1904.43 = 23 mm
Actual eccenY = My / P = 0.72/1904.43 = 0 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(23,22) = 23 mm
eccenY = max(Actual eccenY,eccenYMin) = max(0,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(23 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 18 nos. (3619 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00313 355.01 8.92 346.09 139.17 177 24.63
2 402 0.00278 352.00 8.92 343.08 137.96 133 18.31
3 402 0.00244 343.53 8.92 334.61 134.55 89 11.91
4 402 0.00209 331.20 8.92 322.28 129.60 44 5.73
5 402 0.00175 314.19 8.78 305.41 122.81 0 0.00
6 402 0.00141 281.17 8.13 273.03 109.79 -44 -4.86
7 402 0.00106 212.33 6.96 205.38 82.59 -89 -7.31
8 402 0.00072 143.50 5.25 138.25 55.59 -133 -7.38
9 402 0.00037 74.67 3.02 71.65 28.81 -177 -5.10
Total940.88 35.94
xux = 450 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 450/450 = 0.360
Puc = 0.360 x 20.00 x 300 x 450
= 972.00 kN
Pux1 = Puc + Pus(Total)
= 972.00 + (940.88)
= 1912.88 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 450/450
= 0.416
Muc = 972.00 x (0.5 x 450 - 0.416 x 450) = 36.74 kN-m
Mux1 = Muc + Mus(Total)
= 36.74 + (35.94)
= 72.68 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1810 0.00281 352.22 8.92 343.30 621.21 102 63.36
2 1810 0.00076 152.59 5.51 147.08 266.16 -102 -27.15
Total887.37 36.22
xuy = 328 mm               Puc = C1.fck.B.D
ku = 1.094
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.843
C1 = 0.446 x (1 - C3/6) = 0.383
Puc = 0.383 x 20.00 x 450 x 300
= 1034.93 kN
Puy1 = Puc + Pus(Total)
= 1034.93 + (887.37)
= 1922.30 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.442
Muc = 1034.93 x (0.5 x 300 - 0.442 x 300) = 18.14 kN-m
Muy1 = Muc + Mus(Total)
= 18.14 + (36.22)
= 54.35 kN-m
Pu/Puz = 0.825
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.825, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((43.14/72.68)2.000) + ((38.09/54.35)2.000)
= 0.843 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG18 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 18 nos. (3619 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG19 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3450 mm
From analysis results,loads on column
Axial load,(P) = 481.97 kN
MomentX,(Mx) = 15.99 kN-m
MomentY,(My) = 39.61 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C270.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3450 x 1.000 ) / 450 = 7.67 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3450 x 1.000 ) / 300 = 11.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 10.555 kN.m
My_MinEccen = 9.639 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 15.994 kN.m
My = max(My,My_MinEccen) + MuaddY = 39.608 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 15.99/481.97 = 33 mm
Actual eccenY = My / P = 39.61/481.97 = 82 mm
eccenXMin = (L/500) + (D/30) = (3450/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3450/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(33,22) = 33 mm
eccenY = max(Actual eccenY,eccenYMin) = max(82,20) = 82 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(33 mm) > 0.05 x 300(15 mm)
and eccenY(82 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00278 352.01 8.92 343.09 77.61 179 13.89
2 226 0.00139 278.44 8.10 270.35 61.15 90 5.47
3 226 0.00000 0.00 0.00 0.00 0.00 0 0.00
4 226 -0.00139 -278.44 0.00 -278.44 -62.98 -90 5.64
5 226 -0.00278 -352.01 0.00 -352.01 -79.62 -179 14.25
Total-3.85 39.25
xux = 225 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 225/450 = 0.180
Puc = 0.180 x 20.00 x 300 x 450
= 486.00 kN
Pux1 = Puc + Pus(Total)
= 486.00 + (-3.85)
= 482.15 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 225/450
= 0.208
Muc = 486.00 x (0.5 x 450 - 0.208 x 450) = 63.86 kN-m
Mux1 = Muc + Mus(Total)
= 63.86 + (39.25)
= 103.11 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00243 343.44 8.92 334.52 189.17 104 19.67
2 565 -0.00238 -341.72 0.00 -341.72 -193.24 -104 20.10
Total-4.07 39.77
xuy = 151 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 151/300 = 0.181
Puc = 0.181 x 20.00 x 450 x 300
= 489.80 kN
Puy1 = Puc + Pus(Total)
= 489.80 + (-4.07)
= 485.72 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 151/300
= 0.210
Muc = 489.80 x (0.5 x 300 - 0.210 x 300) = 42.67 kN-m
Muy1 = Muc + Mus(Total)
= 42.67 + (39.77)
= 82.44 kN-m
Pu/Puz = 0.310
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.310, an = 1.183
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((15.99/103.11)1.183) + ((39.61/82.44)1.183)
= 0.531 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG19 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3450 mm
From analysis results,loads on column
Axial load,(P) = 953.50 kN
MomentX,(Mx) = 16.64 kN-m
MomentY,(My) = 36.96 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C270.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3450 x 1.000 ) / 450 = 7.67 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3450 x 1.000 ) / 300 = 11.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 20.882 kN.m
My_MinEccen = 19.070 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 20.882 kN.m
My = max(My,My_MinEccen) + MuaddY = 36.963 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 16.64/953.50 = 17 mm
Actual eccenY = My / P = 36.96/953.50 = 39 mm
eccenXMin = (L/500) + (D/30) = (3450/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3450/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(17,22) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(39,20) = 39 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(39 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00304 354.27 8.92 345.35 78.12 179 13.98
2 226 0.00215 333.29 8.92 324.37 73.37 90 6.57
3 226 0.00126 252.00 7.70 244.30 55.26 0 0.00
4 226 0.00037 73.80 2.99 70.81 16.02 -90 -1.43
5 226 -0.00052 -104.41 0.00 -104.41 -23.62 -179 4.23
Total199.15 23.34
xux = 352 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 352/450 = 0.281
Puc = 0.281 x 20.00 x 300 x 450
= 759.38 kN
Pux1 = Puc + Pus(Total)
= 759.38 + (199.15)
= 958.52 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 352/450
= 0.325
Muc = 759.38 x (0.5 x 450 - 0.325 x 450) = 59.80 kN-m
Mux1 = Muc + Mus(Total)
= 59.80 + (23.34)
= 83.14 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00283 352.44 8.92 343.52 194.26 104 20.20
2 565 -0.00018 -36.52 0.00 -36.52 -20.65 -104 2.15
Total173.61 22.35
xuy = 241 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 241/300 = 0.290
Puc = 0.290 x 20.00 x 450 x 300
= 782.16 kN
Puy1 = Puc + Pus(Total)
= 782.16 + (173.61)
= 955.76 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 241/300
= 0.335
Muc = 782.16 x (0.5 x 300 - 0.335 x 300) = 38.78 kN-m
Muy1 = Muc + Mus(Total)
= 38.78 + (22.35)
= 61.13 kN-m
Pu/Puz = 0.612
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.612, an = 1.687
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((20.88/83.14)1.687) + ((36.96/61.13)1.687)
= 0.525 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG19 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3150 mm
From analysis results,loads on column
Axial load,(P) = 1431.59 kN
MomentX,(Mx) = 10.89 kN-m
MomentY,(My) = 35.99 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C270.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3150 x 1.000 ) / 450 = 7.00 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3150 x 1.000 ) / 300 = 10.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 30.493 kN.m
My_MinEccen = 28.632 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 30.493 kN.m
My = max(My,My_MinEccen) + MuaddY = 35.994 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 10.89/1431.59 = 8 mm
Actual eccenY = My / P = 35.99/1431.59 = 25 mm
eccenXMin = (L/500) + (D/30) = (3150/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (3150/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(8,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(25,20) = 25 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(25 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 6 nos. + #16 - 6 nos. (1885 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00313 355.01 8.92 346.09 139.17 177 24.63
2 226 0.00257 346.99 8.92 338.07 76.47 106 8.09
3 226 0.00203 329.01 8.92 320.09 72.40 37 2.65
4 402 0.00148 292.57 8.32 284.25 114.31 -35 -3.95
5 226 0.00093 185.42 6.35 179.07 40.50 -106 -4.29
6 402 0.00037 74.67 3.02 71.65 28.81 -177 -5.10
Total471.66 22.03
xux = 450 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 450/450 = 0.360
Puc = 0.360 x 20.00 x 300 x 450
= 972.00 kN
Pux1 = Puc + Pus(Total)
= 972.00 + (471.66)
= 1443.66 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 450/450
= 0.416
Muc = 972.00 x (0.5 x 450 - 0.416 x 450) = 36.74 kN-m
Mux1 = Muc + Mus(Total)
= 36.74 + (22.03)
= 58.78 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 942 0.00289 352.95 8.92 344.03 324.24 102 33.07
2 942 0.00063 126.93 4.76 122.17 115.14 -102 -11.74
Total439.38 21.33
xuy = 309 mm               Puc = C1.fck.B.D
ku = 1.031
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.027
C1 = 0.446 x (1 - C3/6) = 0.370
Puc = 0.370 x 20.00 x 450 x 300
= 998.00 kN
Puy1 = Puc + Pus(Total)
= 998.00 + (439.38)
= 1437.38 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.426
Muc = 998.00 x (0.5 x 300 - 0.426 x 300) = 22.09 kN-m
Muy1 = Muc + Mus(Total)
= 22.09 + (21.33)
= 43.42 kN-m
Pu/Puz = 0.802
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.802, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((30.49/58.78)2.000) + ((35.99/43.42)2.000)
= 0.956 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG19 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 6 nos. + #16 - 6 nos. (1885 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3150 mm
From analysis results,loads on column
Axial load,(P) = 2017.70 kN
MomentX,(Mx) = 2.97 kN-m
MomentY,(My) = 34.12 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C270.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3150 x 1.000 ) / 450 = 7.00 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3150 x 1.000 ) / 300 = 10.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 42.977 kN.m
My_MinEccen = 40.354 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 42.977 kN.m
My = max(My,My_MinEccen) + MuaddY = 40.354 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 2.97/2017.70 = 1 mm
Actual eccenY = My / P = 34.12/2017.70 = 17 mm
eccenXMin = (L/500) + (D/30) = (3150/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (3150/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(1,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(17,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 14 nos. (4398 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00310 354.73 8.92 345.81 217.28 175 38.02
2 628 0.00262 348.28 8.92 339.36 213.23 117 24.88
3 628 0.00215 333.29 8.92 324.37 203.81 58 11.89
4 628 0.00168 309.75 8.69 301.06 189.16 0 0.00
5 628 0.00121 241.34 7.52 233.83 146.92 -58 -8.57
6 628 0.00073 146.91 5.35 141.56 88.95 -117 -10.38
7 628 0.00026 52.48 2.19 50.30 31.60 -175 -5.53
Total1090.94 50.31
xux = 432 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 432/450 = 0.346
Puc = 0.346 x 20.00 x 300 x 450
= 934.03 kN
Pux1 = Puc + Pus(Total)
= 934.03 + (1090.94)
= 2024.97 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 432/450
= 0.400
Muc = 934.03 x (0.5 x 450 - 0.400 x 450) = 42.14 kN-m
Mux1 = Muc + Mus(Total)
= 42.14 + (50.31)
= 92.45 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 2199 0.00287 352.75 8.92 343.83 756.13 100 75.61
2 2199 0.00066 131.36 4.90 126.46 278.10 -100 -27.81
Total1034.24 47.80
xuy = 309 mm               Puc = C1.fck.B.D
ku = 1.031
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.027
C1 = 0.446 x (1 - C3/6) = 0.370
Puc = 0.370 x 20.00 x 450 x 300
= 998.00 kN
Puy1 = Puc + Pus(Total)
= 998.00 + (1034.24)
= 2032.23 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.426
Muc = 998.00 x (0.5 x 300 - 0.426 x 300) = 22.09 kN-m
Muy1 = Muc + Mus(Total)
= 22.09 + (47.80)
= 69.90 kN-m
Pu/Puz = 0.793
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.793, an = 1.988
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((42.98/92.45)1.988) + ((40.35/69.90)1.988)
= 0.554 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 20 = 320 mm
               iii) 300 mm
               Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG19 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 14 nos. (4398 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3675 mm
From analysis results,loads on column
Axial load,(P) = 2582.77 kN
MomentX,(Mx) = 4.30 kN-m
MomentY,(My) = 21.22 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C270.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3675 x 1.000 ) / 450 = 8.17 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3675 x 1.000 ) / 300 = 12.25 > 12.00,
column is slender in this direction.
Puz = (0.45 x fck x Ag) + (0.75 x fy - 0.45 x fck) x Ast
= 0.45 x 20.00 x 135000.01 + (0.75 x 415.00 - 0.45 x 20.00 ) x 6872.23 = 3292132.813 N
k1 = 0.190000
k2 = 0.012000 MPa
Pub = (k1 x fck x Ag) + (k2 x Ast x 100)
= (0.190000 x 20.00 x 135000.01) + (0.01 x 6872.23 x 100) = 521246.704 N
k = (Puz - Pu) / (Puz - Pub) <= 1
= (3292132.813 - 2582771.973) / (3292132.813 - 521246.704) = 0.256005
MuaddY = Pu x (Width/2000) x (Lo/Width) x (Lo/Width) x k
= 2582771.97 x (300.00/2000) x (3675.00/300.00) x (3675.00/300.00) x 0.256005
= 14883259.773 N.mm = 14.883 kN.m
Mx_MinEccen = 57.725 kN.m
My_MinEccen = 51.655 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 57.725 kN.m
My = max(My,My_MinEccen) + MuaddY = 66.539 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 4.30/2582.77 = 2 mm
Actual eccenY = My / P = 21.22/2582.77 = 8 mm
eccenXMin = (L/500) + (D/30) = (3675/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3675/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(2,22) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(8,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #25 - 14 nos. (6872 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 982 0.00307 354.50 8.92 345.58 339.27 173 58.52
2 982 0.00259 347.56 8.92 338.64 332.45 115 38.23
3 982 0.00212 332.22 8.92 323.30 317.39 58 18.25
4 982 0.00165 307.87 8.64 299.23 293.76 0 0.00
5 982 0.00118 235.13 7.40 227.73 223.57 -58 -12.86
6 982 0.00070 140.51 5.17 135.35 132.88 -115 -15.28
7 982 0.00023 45.90 1.93 43.97 43.16 -173 -7.45
Total1682.49 79.42
xux = 425 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 425/450 = 0.340
Puc = 0.340 x 20.00 x 300 x 450
= 918.84 kN
Pux1 = Puc + Pus(Total)
= 918.84 + (1682.49)
= 2601.34 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 425/450
= 0.393
Muc = 918.84 x (0.5 x 450 - 0.393 x 450) = 44.14 kN-m
Mux1 = Muc + Mus(Total)
= 44.14 + (79.42)
= 123.56 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 3436 0.00286 352.71 8.92 343.79 1181.30 98 115.18
2 3436 0.00065 129.89 4.85 125.03 429.63 -98 -41.89
Total1610.93 73.29
xuy = 305 mm               Puc = C1.fck.B.D
ku = 1.016
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.083
C1 = 0.446 x (1 - C3/6) = 0.366
Puc = 0.366 x 20.00 x 450 x 300
= 986.88 kN
Puy1 = Puc + Pus(Total)
= 986.88 + (1610.93)
= 2597.80 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.421
Muc = 986.88 x (0.5 x 300 - 0.421 x 300) = 23.28 kN-m
Muy1 = Muc + Mus(Total)
= 23.28 + (73.29)
= 96.57 kN-m
Pu/Puz = 0.785
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.785, an = 1.974
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((57.72/123.56)1.974) + ((66.54/96.57)1.974)
= 0.702 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 25/4 = 6 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (6, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 25 = 400 mm
               iii) 300 mm
               Required spacing = minimum of (300, 400, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG19 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #25 - 14 nos. (6872 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG20 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 367.06 kN
MomentX,(Mx) = 13.88 kN-m
MomentY,(My) = 78.79 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C280.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 7.598 kN.m
My_MinEccen = 7.341 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 13.876 kN.m
My = max(My,My_MinEccen) + MuaddY = 78.791 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 13.88/367.06 = 38 mm
Actual eccenY = My / P = 78.79/367.06 = 215 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(38,21) = 38 mm
eccenY = max(Actual eccenY,eccenYMin) = max(215,20) = 215 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(38 mm) > 0.05 x 300(15 mm)
and eccenY(215 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 2 nos. + #16 - 6 nos. (1433 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00269 350.00 8.92 341.08 137.16 177 24.28
2 226 0.00071 142.04 5.21 136.83 30.95 60 1.85
3 402 -0.00127 -253.93 0.00 -253.93 -102.11 -58 5.89
4 402 -0.00328 -356.38 0.00 -356.38 -143.31 -177 25.37
Total-77.31 57.38
xux = 207 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 207/450 = 0.166
Puc = 0.166 x 20.00 x 300 x 450
= 448.03 kN
Pux1 = Puc + Pus(Total)
= 448.03 + (-77.31)
= 370.72 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 207/450
= 0.192
Muc = 448.03 x (0.5 x 450 - 0.192 x 450) = 62.15 kN-m
Mux1 = Muc + Mus(Total)
= 62.15 + (57.38)
= 119.53 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 716 0.00212 332.20 8.92 323.28 231.56 102 23.62
2 716 -0.00374 -360.35 0.00 -360.35 -258.11 -102 26.33
Total-26.55 49.95
xuy = 122 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 122/300 = 0.146
Puc = 0.146 x 20.00 x 450 x 300
= 394.88 kN
Puy1 = Puc + Pus(Total)
= 394.88 + (-26.55)
= 368.33 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 122/300
= 0.169
Muc = 394.88 x (0.5 x 300 - 0.169 x 300) = 39.21 kN-m
Muy1 = Muc + Mus(Total)
= 39.21 + (49.95)
= 89.16 kN-m
Pu/Puz = 0.223
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.223, an = 1.038
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((13.88/119.53)1.038) + ((78.79/89.16)1.038)
= 0.987 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG20 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 2 nos. + #16 - 6 nos. (1433 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 750.68 kN
MomentX,(Mx) = 12.76 kN-m
MomentY,(My) = 71.75 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C280.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 15.539 kN.m
My_MinEccen = 15.014 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 15.539 kN.m
My = max(My,My_MinEccen) + MuaddY = 71.749 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 12.76/750.68 = 17 mm
Actual eccenY = My / P = 71.75/750.68 = 96 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(17,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(96,20) = 96 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(96 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 4 nos. + #16 - 4 nos. (1257 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00293 353.33 8.92 344.41 138.49 177 24.51
2 226 0.00154 297.83 8.44 289.39 65.46 58 3.82
3 226 0.00016 32.37 1.39 30.98 7.01 -58 -0.41
4 402 -0.00124 -247.25 0.00 -247.25 -99.43 -177 17.60
Total111.54 45.52
xux = 297 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 297/450 = 0.238
Puc = 0.238 x 20.00 x 300 x 450
= 641.67 kN
Pux1 = Puc + Pus(Total)
= 641.67 + (111.54)
= 753.21 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 297/450
= 0.275
Muc = 641.67 x (0.5 x 450 - 0.275 x 450) = 65.08 kN-m
Mux1 = Muc + Mus(Total)
= 65.08 + (45.52)
= 110.60 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00267 349.40 8.92 340.48 213.93 102 21.82
2 628 -0.00088 -175.16 0.00 -175.16 -110.06 -102 11.23
Total103.87 33.05
xuy = 202 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 202/300 = 0.242
Puc = 0.242 x 20.00 x 450 x 300
= 653.06 kN
Puy1 = Puc + Pus(Total)
= 653.06 + (103.87)
= 756.93 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 202/300
= 0.280
Muc = 653.06 x (0.5 x 300 - 0.280 x 300) = 43.20 kN-m
Muy1 = Muc + Mus(Total)
= 43.20 + (33.05)
= 76.25 kN-m
Pu/Puz = 0.471
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.471, an = 1.451
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((15.54/110.60)1.451) + ((71.75/76.25)1.451)
= 0.974 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG20 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 4 nos. + #16 - 4 nos. (1257 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1129.64 kN
MomentX,(Mx) = 11.37 kN-m
MomentY,(My) = 69.20 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C280.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 23.384 kN.m
My_MinEccen = 22.593 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 23.384 kN.m
My = max(My,My_MinEccen) + MuaddY = 69.201 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 11.37/1129.64 = 10 mm
Actual eccenY = My / P = 69.20/1129.64 = 61 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(10,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(61,20) = 61 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(61 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 18 nos. (2036 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00304 354.27 8.92 345.35 78.12 179 13.98
2 226 0.00260 347.60 8.92 338.68 76.61 134 10.28
3 226 0.00215 333.29 8.92 324.37 73.37 90 6.57
4 226 0.00171 311.41 8.73 302.69 68.47 45 3.06
5 226 0.00126 252.00 7.70 244.30 55.26 0 0.00
6 226 0.00081 162.90 5.79 157.11 35.54 -45 -1.59
7 226 0.00037 73.80 2.99 70.81 16.02 -90 -1.43
8 226 -0.00008 -15.31 0.00 -15.31 -3.46 -134 0.46
9 226 -0.00052 -104.41 0.00 -104.41 -23.62 -179 4.23
Total376.29 35.57
xux = 352 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 352/450 = 0.281
Puc = 0.281 x 20.00 x 300 x 450
= 759.38 kN
Pux1 = Puc + Pus(Total)
= 759.38 + (376.29)
= 1135.67 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 352/450
= 0.325
Muc = 759.38 x (0.5 x 450 - 0.325 x 450) = 59.80 kN-m
Mux1 = Muc + Mus(Total)
= 59.80 + (35.57)
= 95.37 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1018 0.00285 352.60 8.92 343.68 349.83 104 36.38
2 1018 -0.00008 -15.67 0.00 -15.67 -15.95 -104 1.66
Total333.88 38.04
xuy = 248 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 248/300 = 0.298
Puc = 0.298 x 20.00 x 450 x 300
= 804.94 kN
Puy1 = Puc + Pus(Total)
= 804.94 + (333.88)
= 1138.81 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 248/300
= 0.345
Muc = 804.94 x (0.5 x 300 - 0.345 x 300) = 37.55 kN-m
Muy1 = Muc + Mus(Total)
= 37.55 + (38.04)
= 75.59 kN-m
Pu/Puz = 0.617
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.617, an = 1.695
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((23.38/95.37)1.695) + ((69.20/75.59)1.695)
= 0.953 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG20 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 18 nos. (2036 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1501.51 kN
MomentX,(Mx) = 9.04 kN-m
MomentY,(My) = 65.41 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C280.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 31.081 kN.m
My_MinEccen = 30.030 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 31.081 kN.m
My = max(My,My_MinEccen) + MuaddY = 65.406 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 9.04/1501.51 = 6 mm
Actual eccenY = My / P = 65.41/1501.51 = 44 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(6,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(44,20) = 44 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(44 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 16 nos. (3217 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00306 354.44 8.92 345.52 138.94 177 24.59
2 402 0.00260 347.68 8.92 338.76 136.22 126 17.22
3 402 0.00214 332.80 8.92 323.88 130.24 76 9.88
4 402 0.00168 309.56 8.69 300.88 120.99 25 3.06
5 402 0.00121 242.80 7.54 235.26 94.60 -25 -2.39
6 402 0.00075 150.42 5.45 144.97 58.30 -76 -4.42
7 402 0.00029 58.04 2.40 55.64 22.37 -126 -2.83
8 402 -0.00017 -34.34 0.00 -34.34 -13.81 -177 2.44
Total687.86 47.55
xux = 383 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 383/450 = 0.307
Puc = 0.307 x 20.00 x 300 x 450
= 827.72 kN
Pux1 = Puc + Pus(Total)
= 827.72 + (687.86)
= 1515.58 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 383/450
= 0.354
Muc = 827.72 x (0.5 x 450 - 0.354 x 450) = 54.29 kN-m
Mux1 = Muc + Mus(Total)
= 54.29 + (47.55)
= 101.84 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1608 0.00288 352.87 8.92 343.95 553.24 102 56.43
2 1608 0.00026 51.17 2.14 49.04 78.87 -102 -8.05
Total632.11 48.39
xuy = 272 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 272/300 = 0.326
Puc = 0.326 x 20.00 x 450 x 300
= 880.88 kN
Puy1 = Puc + Pus(Total)
= 880.88 + (632.11)
= 1512.99 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 272/300
= 0.377
Muc = 880.88 x (0.5 x 300 - 0.377 x 300) = 32.50 kN-m
Muy1 = Muc + Mus(Total)
= 32.50 + (48.39)
= 80.89 kN-m
Pu/Puz = 0.686
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.686, an = 1.811
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((31.08/101.84)1.811) + ((65.41/80.89)1.811)
= 0.797 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG20 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 16 nos. (3217 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 1858.99 kN
MomentX,(Mx) = 5.33 kN-m
MomentY,(My) = 42.84 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C280.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 40.433 kN.m
My_MinEccen = 37.180 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 40.433 kN.m
My = max(My,My_MinEccen) + MuaddY = 42.838 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 5.33/1858.99 = 3 mm
Actual eccenY = My / P = 42.84/1858.99 = 23 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(3,22) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(23,20) = 23 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(23 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 18 nos. (3619 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00312 354.93 8.92 346.01 139.14 177 24.63
2 402 0.00277 351.85 8.92 342.93 137.90 133 18.31
3 402 0.00241 342.87 8.92 333.95 134.29 89 11.88
4 402 0.00206 329.96 8.92 321.04 129.10 44 5.71
5 402 0.00171 311.57 8.73 302.84 121.78 0 0.00
6 402 0.00136 271.11 7.99 263.12 105.81 -44 -4.68
7 402 0.00100 200.63 6.70 193.93 77.98 -89 -6.90
8 402 0.00065 130.14 4.86 125.28 50.38 -133 -6.69
9 402 0.00030 59.66 2.46 57.20 23.00 -177 -4.07
Total919.37 38.19
xux = 439 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 439/450 = 0.352
Puc = 0.352 x 20.00 x 300 x 450
= 949.22 kN
Pux1 = Puc + Pus(Total)
= 949.22 + (919.37)
= 1868.59 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 439/450
= 0.406
Muc = 949.22 x (0.5 x 450 - 0.406 x 450) = 40.05 kN-m
Mux1 = Muc + Mus(Total)
= 40.05 + (38.19)
= 78.23 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1810 0.00287 352.75 8.92 343.83 622.18 102 63.46
2 1810 0.00067 133.83 4.97 128.86 233.19 -102 -23.78
Total855.37 39.68
xuy = 314 mm               Puc = C1.fck.B.D
ku = 1.047
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.976
C1 = 0.446 x (1 - C3/6) = 0.373
Puc = 0.373 x 20.00 x 450 x 300
= 1008.29 kN
Puy1 = Puc + Pus(Total)
= 1008.29 + (855.37)
= 1863.66 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.431
Muc = 1008.29 x (0.5 x 300 - 0.431 x 300) = 20.99 kN-m
Muy1 = Muc + Mus(Total)
= 20.99 + (39.68)
= 60.67 kN-m
Pu/Puz = 0.805
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.805, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((40.43/78.23)2.000) + ((42.84/60.67)2.000)
= 0.766 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG20 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 18 nos. (3619 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG21 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 292.51 kN
MomentX,(Mx) = 9.35 kN-m
MomentY,(My) = 60.85 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C290.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 6.055 kN.m
My_MinEccen = 5.850 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 9.346 kN.m
My = max(My,My_MinEccen) + MuaddY = 60.854 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 9.35/292.51 = 32 mm
Actual eccenY = My / P = 60.85/292.51 = 208 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(32,21) = 32 mm
eccenY = max(Actual eccenY,eccenYMin) = max(208,20) = 208 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(32 mm) > 0.05 x 300(15 mm)
and eccenY(208 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00259 347.51 8.92 338.59 76.59 179 13.71
2 226 0.00083 165.75 5.86 159.89 36.17 90 3.24
3 226 -0.00094 -187.13 0.00 -187.13 -42.33 0 0.00
4 226 -0.00270 -350.26 0.00 -350.26 -79.23 -90 7.09
5 226 -0.00446 -360.90 0.00 -360.90 -81.63 -179 14.61
Total-90.43 38.65
xux = 178 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 178/450 = 0.142
Puc = 0.142 x 20.00 x 300 x 450
= 383.48 kN
Pux1 = Puc + Pus(Total)
= 383.48 + (-90.43)
= 293.05 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 178/450
= 0.164
Muc = 383.48 x (0.5 x 450 - 0.164 x 450) = 57.96 kN-m
Mux1 = Muc + Mus(Total)
= 57.96 + (38.65)
= 96.61 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00187 321.94 8.88 313.05 177.03 104 18.41
2 565 -0.00548 -360.90 0.00 -360.90 -204.08 -104 21.22
Total-27.06 39.64
xuy = 99 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 99/300 = 0.119
Puc = 0.119 x 20.00 x 450 x 300
= 320.84 kN
Puy1 = Puc + Pus(Total)
= 320.84 + (-27.06)
= 293.78 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 99/300
= 0.137
Muc = 320.84 x (0.5 x 300 - 0.137 x 300) = 34.91 kN-m
Muy1 = Muc + Mus(Total)
= 34.91 + (39.64)
= 74.54 kN-m
Pu/Puz = 0.188
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.188, an = 1.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((9.35/96.61)1.000) + ((60.85/74.54)1.000)
= 0.913 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG21 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 607.56 kN
MomentX,(Mx) = 8.66 kN-m
MomentY,(My) = 55.68 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C290.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 12.576 kN.m
My_MinEccen = 12.151 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 12.576 kN.m
My = max(My,My_MinEccen) + MuaddY = 55.678 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 8.66/607.56 = 14 mm
Actual eccenY = My / P = 55.68/607.56 = 92 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(14,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(92,20) = 92 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(92 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00287 352.79 8.92 343.87 77.78 179 13.92
2 226 0.00165 308.08 8.65 299.43 67.73 90 6.06
3 226 0.00043 86.30 3.43 82.87 18.74 0 0.00
4 226 -0.00079 -157.81 0.00 -157.81 -35.70 -90 3.19
5 226 -0.00201 -328.09 0.00 -328.09 -74.21 -179 13.28
Total54.34 36.46
xux = 257 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 257/450 = 0.205
Puc = 0.205 x 20.00 x 300 x 450
= 554.34 kN
Pux1 = Puc + Pus(Total)
= 554.34 + (54.34)
= 608.69 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 257/450
= 0.237
Muc = 554.34 x (0.5 x 450 - 0.237 x 450) = 65.54 kN-m
Mux1 = Muc + Mus(Total)
= 65.54 + (36.46)
= 102.01 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00260 347.74 8.92 338.82 191.60 104 19.93
2 565 -0.00146 -290.43 0.00 -290.43 -164.23 -104 17.08
Total27.36 37.01
xuy = 179 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 179/300 = 0.215
Puc = 0.215 x 20.00 x 450 x 300
= 580.92 kN
Puy1 = Puc + Pus(Total)
= 580.92 + (27.36)
= 608.29 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 179/300
= 0.249
Muc = 580.92 x (0.5 x 300 - 0.249 x 300) = 43.81 kN-m
Muy1 = Muc + Mus(Total)
= 43.81 + (37.01)
= 80.82 kN-m
Pu/Puz = 0.390
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.390, an = 1.317
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((12.58/102.01)1.317) + ((55.68/80.82)1.317)
= 0.676 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG21 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 913.38 kN
MomentX,(Mx) = 8.11 kN-m
MomentY,(My) = 50.80 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C290.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 18.907 kN.m
My_MinEccen = 18.268 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 18.907 kN.m
My = max(My,My_MinEccen) + MuaddY = 50.801 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 8.11/913.38 = 9 mm
Actual eccenY = My / P = 50.80/913.38 = 56 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(9,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(56,20) = 56 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(56 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00303 354.14 8.92 345.22 78.09 179 13.98
2 226 0.00211 331.75 8.92 322.83 73.02 90 6.54
3 226 0.00119 238.14 7.46 230.68 52.18 0 0.00
4 226 0.00027 54.43 2.26 52.17 11.80 -90 -1.06
5 226 -0.00065 -129.29 0.00 -129.29 -29.24 -179 5.23
Total185.85 24.69
xux = 341 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 341/450 = 0.273
Puc = 0.273 x 20.00 x 300 x 450
= 736.59 kN
Pux1 = Puc + Pus(Total)
= 736.59 + (185.85)
= 922.44 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 341/450
= 0.315
Muc = 736.59 x (0.5 x 450 - 0.315 x 450) = 61.24 kN-m
Mux1 = Muc + Mus(Total)
= 61.24 + (24.69)
= 85.93 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00281 352.26 8.92 343.34 194.16 104 20.19
2 565 -0.00029 -58.61 0.00 -58.61 -33.15 -104 3.45
Total161.01 23.64
xuy = 234 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 234/300 = 0.281
Puc = 0.281 x 20.00 x 450 x 300
= 759.38 kN
Puy1 = Puc + Pus(Total)
= 759.38 + (161.01)
= 920.39 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 234/300
= 0.325
Muc = 759.38 x (0.5 x 300 - 0.325 x 300) = 39.87 kN-m
Muy1 = Muc + Mus(Total)
= 39.87 + (23.64)
= 63.51 kN-m
Pu/Puz = 0.587
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.587, an = 1.644
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((18.91/85.93)1.644) + ((50.80/63.51)1.644)
= 0.776 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG21 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1206.16 kN
MomentX,(Mx) = 7.14 kN-m
MomentY,(My) = 43.68 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C290.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 24.967 kN.m
My_MinEccen = 24.123 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 24.967 kN.m
My = max(My,My_MinEccen) + MuaddY = 43.677 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 7.14/1206.16 = 6 mm
Actual eccenY = My / P = 43.68/1206.16 = 36 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(6,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(36,20) = 36 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(36 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 2 nos. + #16 - 6 nos. (1433 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00310 354.76 8.92 345.84 139.07 177 24.62
2 226 0.00212 332.03 8.92 323.11 73.09 60 4.36
3 402 0.00114 227.04 7.25 219.79 88.38 -58 -5.10
4 402 0.00014 27.37 1.18 26.19 10.53 -177 -1.86
Total311.07 22.02
xux = 418 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 418/450 = 0.335
Puc = 0.335 x 20.00 x 300 x 450
= 903.66 kN
Pux1 = Puc + Pus(Total)
= 903.66 + (311.07)
= 1214.73 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 418/450
= 0.387
Muc = 903.66 x (0.5 x 450 - 0.387 x 450) = 46.05 kN-m
Mux1 = Muc + Mus(Total)
= 46.05 + (22.02)
= 68.07 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 716 0.00290 353.05 8.92 344.13 246.49 102 25.14
2 716 0.00036 72.80 2.95 69.85 50.03 -102 -5.10
Total296.52 20.04
xuy = 281 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 281/300 = 0.338
Puc = 0.338 x 20.00 x 450 x 300
= 911.25 kN
Puy1 = Puc + Pus(Total)
= 911.25 + (296.52)
= 1207.77 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 281/300
= 0.390
Muc = 911.25 x (0.5 x 300 - 0.390 x 300) = 30.07 kN-m
Muy1 = Muc + Mus(Total)
= 30.07 + (20.04)
= 50.11 kN-m
Pu/Puz = 0.732
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.732, an = 1.886
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((24.97/68.07)1.886) + ((43.68/50.11)1.886)
= 0.922 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG21 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 2 nos. + #16 - 6 nos. (1433 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 1480.17 kN
MomentX,(Mx) = 4.52 kN-m
MomentY,(My) = 25.90 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C290.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 32.194 kN.m
My_MinEccen = 29.603 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 32.194 kN.m
My = max(My,My_MinEccen) + MuaddY = 29.603 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 4.52/1480.17 = 3 mm
Actual eccenY = My / P = 25.90/1480.17 = 17 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(3,22) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(17,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 6 nos. + #16 - 6 nos. (1885 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00304 354.26 8.92 345.34 138.87 177 24.58
2 226 0.00253 345.87 8.92 336.95 76.22 106 8.06
3 226 0.00203 328.92 8.92 320.00 72.38 37 2.65
4 402 0.00152 296.30 8.41 287.90 115.77 -35 -4.01
5 226 0.00101 201.69 6.73 194.96 44.10 -106 -4.67
6 402 0.00050 99.33 3.88 95.45 38.38 -177 -6.79
Total485.72 19.83
xux = 471 mm               Puc = C1.fck.B.D
ku = 1.047
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.976
C1 = 0.446 x (1 - C3/6) = 0.373
Puc = 0.373 x 20.00 x 300 x 450
= 1008.29 kN
Pux1 = Puc + Pus(Total)
= 1008.29 + (485.72)
= 1494.01 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.431
Muc = 1008.29 x (0.5 x 450 - 0.431 x 450) = 31.49 kN-m
Mux1 = Muc + Mus(Total)
= 31.49 + (19.83)
= 51.31 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 942 0.00283 352.39 8.92 343.47 323.71 102 33.02
2 942 0.00073 146.64 5.34 141.30 133.17 -102 -13.58
Total456.88 19.43
xuy = 323 mm               Puc = C1.fck.B.D
ku = 1.078
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.884
C1 = 0.446 x (1 - C3/6) = 0.380
Puc = 0.380 x 20.00 x 450 x 300
= 1026.69 kN
Puy1 = Puc + Pus(Total)
= 1026.69 + (456.88)
= 1483.56 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.438
Muc = 1026.69 x (0.5 x 300 - 0.438 x 300) = 19.02 kN-m
Muy1 = Muc + Mus(Total)
= 19.02 + (19.43)
= 38.45 kN-m
Pu/Puz = 0.829
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.829, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((32.19/51.31)2.000) + ((29.60/38.45)2.000)
= 0.986 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG21 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 6 nos. + #16 - 6 nos. (1885 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG22 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 697.38 kN
MomentX,(Mx) = 47.01 kN-m
MomentY,(My) = 22.21 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C300.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 14.436 kN.m
My_MinEccen = 13.948 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 47.012 kN.m
My = max(My,My_MinEccen) + MuaddY = 22.210 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 47.01/697.38 = 67 mm
Actual eccenY = My / P = 22.21/697.38 = 32 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(67,21) = 67 mm
eccenY = max(Actual eccenY,eccenYMin) = max(32,20) = 32 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(67 mm) > 0.05 x 300(15 mm)
and eccenY(32 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00293 353.27 8.92 344.35 77.89 179 13.94
2 226 0.00181 318.17 8.84 309.33 69.97 90 6.26
3 226 0.00070 140.00 5.15 134.85 30.50 0 0.00
4 226 -0.00041 -82.76 0.00 -82.76 -18.72 -90 1.68
5 226 -0.00153 -296.99 0.00 -296.99 -67.18 -179 12.02
Total92.46 33.90
xux = 281 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 281/450 = 0.225
Puc = 0.225 x 20.00 x 300 x 450
= 607.50 kN
Pux1 = Puc + Pus(Total)
= 607.50 + (92.46)
= 699.96 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 281/450
= 0.260
Muc = 607.50 x (0.5 x 450 - 0.260 x 450) = 65.61 kN-m
Mux1 = Muc + Mus(Total)
= 65.61 + (33.90)
= 99.51 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00267 349.55 8.92 340.63 192.62 104 20.03
2 565 -0.00107 -213.99 0.00 -213.99 -121.01 -104 12.59
Total71.61 32.62
xuy = 195 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 195/300 = 0.233
Puc = 0.233 x 20.00 x 450 x 300
= 630.28 kN
Puy1 = Puc + Pus(Total)
= 630.28 + (71.61)
= 701.89 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 195/300
= 0.270
Muc = 630.28 x (0.5 x 300 - 0.270 x 300) = 43.54 kN-m
Muy1 = Muc + Mus(Total)
= 43.54 + (32.62)
= 76.15 kN-m
Pu/Puz = 0.448
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.448, an = 1.413
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((47.01/99.51)1.413) + ((22.21/76.15)1.413)
= 0.522 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG22 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1398.21 kN
MomentX,(Mx) = 44.93 kN-m
MomentY,(My) = 21.81 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C300.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 28.943 kN.m
My_MinEccen = 27.964 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 44.930 kN.m
My = max(My,My_MinEccen) + MuaddY = 27.964 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 44.93/1398.21 = 32 mm
Actual eccenY = My / P = 21.81/1398.21 = 16 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(32,21) = 32 mm
eccenY = max(Actual eccenY,eccenYMin) = max(16,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(32 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 16 nos. (1810 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00313 355.07 8.92 346.15 78.30 179 14.02
2 226 0.00273 350.93 8.92 342.01 77.36 128 9.89
3 226 0.00232 339.46 8.92 330.54 74.77 77 5.74
4 226 0.00191 324.28 8.90 315.38 71.34 26 1.82
5 226 0.00150 294.80 8.37 286.42 64.79 -26 -1.66
6 226 0.00110 219.40 7.10 212.30 48.02 -77 -3.68
7 226 0.00069 137.94 5.09 132.85 30.05 -128 -3.84
8 226 0.00028 56.47 2.34 54.13 12.24 -179 -2.19
Total456.86 20.09
xux = 439 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 439/450 = 0.352
Puc = 0.352 x 20.00 x 300 x 450
= 949.22 kN
Pux1 = Puc + Pus(Total)
= 949.22 + (456.86)
= 1406.08 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 439/450
= 0.406
Muc = 949.22 x (0.5 x 450 - 0.406 x 450) = 40.05 kN-m
Mux1 = Muc + Mus(Total)
= 40.05 + (20.09)
= 60.14 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 905 0.00294 353.35 8.92 344.43 311.64 104 32.41
2 905 0.00058 115.12 4.40 110.73 100.18 -104 -10.42
Total411.82 21.99
xuy = 305 mm               Puc = C1.fck.B.D
ku = 1.016
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.083
C1 = 0.446 x (1 - C3/6) = 0.366
Puc = 0.366 x 20.00 x 450 x 300
= 986.88 kN
Puy1 = Puc + Pus(Total)
= 986.88 + (411.82)
= 1398.70 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.421
Muc = 986.88 x (0.5 x 300 - 0.421 x 300) = 23.28 kN-m
Muy1 = Muc + Mus(Total)
= 23.28 + (21.99)
= 45.28 kN-m
Pu/Puz = 0.794
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.794, an = 1.989
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((44.93/60.14)1.989) + ((27.96/45.28)1.989)
= 0.943 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG22 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 16 nos. (1810 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 2099.93 kN
MomentX,(Mx) = 41.20 kN-m
MomentY,(My) = 20.99 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C300.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 43.469 kN.m
My_MinEccen = 41.999 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 43.469 kN.m
My = max(My,My_MinEccen) + MuaddY = 41.999 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 41.20/2099.93 = 20 mm
Actual eccenY = My / P = 20.99/2099.93 = 10 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(20,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(10,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 14 nos. (4398 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00311 354.87 8.92 345.95 217.37 175 38.04
2 628 0.00266 349.16 8.92 340.24 213.78 117 24.94
3 628 0.00220 335.22 8.92 326.30 205.02 58 11.96
4 628 0.00175 314.19 8.78 305.41 191.89 0 0.00
5 628 0.00130 259.26 7.82 251.44 157.99 -58 -9.22
6 628 0.00084 168.52 5.93 162.59 102.16 -117 -11.92
7 628 0.00039 77.78 3.13 74.65 46.90 -175 -8.21
Total1135.11 45.60
xux = 450 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 450/450 = 0.360
Puc = 0.360 x 20.00 x 300 x 450
= 972.00 kN
Pux1 = Puc + Pus(Total)
= 972.00 + (1135.11)
= 2107.11 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 450/450
= 0.416
Muc = 972.00 x (0.5 x 450 - 0.416 x 450) = 36.74 kN-m
Mux1 = Muc + Mus(Total)
= 36.74 + (45.60)
= 82.34 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 2199 0.00281 352.21 8.92 343.29 754.93 100 75.49
2 2199 0.00075 150.74 5.46 145.29 319.51 -100 -31.95
Total1074.43 43.54
xuy = 323 mm               Puc = C1.fck.B.D
ku = 1.078
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.884
C1 = 0.446 x (1 - C3/6) = 0.380
Puc = 0.380 x 20.00 x 450 x 300
= 1026.69 kN
Puy1 = Puc + Pus(Total)
= 1026.69 + (1074.43)
= 2101.12 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.438
Muc = 1026.69 x (0.5 x 300 - 0.438 x 300) = 19.02 kN-m
Muy1 = Muc + Mus(Total)
= 19.02 + (43.54)
= 62.56 kN-m
Pu/Puz = 0.825
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.825, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((43.47/82.34)2.000) + ((42.00/62.56)2.000)
= 0.729 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 20 = 320 mm
               iii) 300 mm
               Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG22 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 14 nos. (4398 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 2806.82 kN
MomentX,(Mx) = 33.44 kN-m
MomentY,(My) = 19.94 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C300.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 58.101 kN.m
My_MinEccen = 56.136 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 58.101 kN.m
My = max(My,My_MinEccen) + MuaddY = 56.136 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 33.44/2806.82 = 12 mm
Actual eccenY = My / P = 19.94/2806.82 = 7 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(12,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(7,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #25 - 14 nos. (6872 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 982 0.00304 354.21 8.92 345.29 338.98 173 58.47
2 982 0.00261 347.97 8.92 339.05 332.86 115 38.28
3 982 0.00219 334.61 8.92 325.69 319.74 58 18.39
4 982 0.00176 315.00 8.79 306.20 300.62 0 0.00
5 982 0.00134 267.79 7.95 259.84 255.10 -58 -14.67
6 982 0.00091 182.98 6.29 176.69 173.46 -115 -19.95
7 982 0.00049 98.17 3.84 94.33 92.61 -173 -15.98
Total1813.37 64.55
xux = 464 mm               Puc = C1.fck.B.D
ku = 1.031
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.027
C1 = 0.446 x (1 - C3/6) = 0.370
Puc = 0.370 x 20.00 x 300 x 450
= 998.00 kN
Pux1 = Puc + Pus(Total)
= 998.00 + (1813.37)
= 2811.37 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.426
Muc = 998.00 x (0.5 x 450 - 0.426 x 450) = 33.14 kN-m
Mux1 = Muc + Mus(Total)
= 33.14 + (64.55)
= 97.69 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 3436 0.00271 350.57 8.92 341.65 1173.95 98 114.46
2 3436 0.00089 177.30 6.16 171.15 588.09 -98 -57.34
Total1762.04 57.12
xuy = 342 mm               Puc = C1.fck.B.D
ku = 1.141
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.736
C1 = 0.446 x (1 - C3/6) = 0.391
Puc = 0.391 x 20.00 x 450 x 300
= 1056.48 kN
Puy1 = Puc + Pus(Total)
= 1056.48 + (1762.04)
= 2818.52 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.450
Muc = 1056.48 x (0.5 x 300 - 0.450 x 300) = 15.83 kN-m
Muy1 = Muc + Mus(Total)
= 15.83 + (57.12)
= 72.95 kN-m
Pu/Puz = 0.853
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.853, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((58.10/97.69)2.000) + ((56.14/72.95)2.000)
= 0.946 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 25/4 = 6 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (6, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 25 = 400 mm
               iii) 300 mm
               Required spacing = minimum of (300, 400, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG22 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #25 - 14 nos. (6872 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 3512.54 kN
MomentX,(Mx) = 18.70 kN-m
MomentY,(My) = 12.95 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C300.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 76.398 kN.m
My_MinEccen = 70.251 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 76.398 kN.m
My = max(My,My_MinEccen) + MuaddY = 70.251 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 18.70/3512.54 = 5 mm
Actual eccenY = My / P = 12.95/3512.54 = 4 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(5,22) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(4,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #40 - 8 nos. (10053 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 2513 0.00303 354.19 8.92 345.27 867.76 165 143.18
2 2513 0.00218 334.27 8.92 325.35 817.69 55 44.97
3 2513 0.00132 264.44 7.90 256.55 644.78 -55 -35.46
4 2513 0.00047 93.33 3.68 89.66 225.33 -165 -37.18
Total2555.56 115.51
xux = 450 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 450/450 = 0.360
Puc = 0.360 x 20.00 x 300 x 450
= 972.00 kN
Pux1 = Puc + Pus(Total)
= 972.00 + (2555.56)
= 3527.56 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 450/450
= 0.416
Muc = 972.00 x (0.5 x 450 - 0.416 x 450) = 36.74 kN-m
Mux1 = Muc + Mus(Total)
= 36.74 + (115.51)
= 152.25 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 5027 0.00272 350.80 8.92 341.88 1718.48 90 154.66
2 5027 0.00083 165.63 5.86 159.78 803.12 -90 -72.28
Total2521.60 82.38
xuy = 319 mm               Puc = C1.fck.B.D
ku = 1.063
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.929
C1 = 0.446 x (1 - C3/6) = 0.377
Puc = 0.377 x 20.00 x 450 x 300
= 1017.83 kN
Puy1 = Puc + Pus(Total)
= 1017.83 + (2521.60)
= 3539.43 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.435
Muc = 1017.83 x (0.5 x 300 - 0.435 x 300) = 19.97 kN-m
Muy1 = Muc + Mus(Total)
= 19.97 + (82.38)
= 102.35 kN-m
Pu/Puz = 0.826
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.826, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((76.40/152.25)2.000) + ((70.25/102.35)2.000)
= 0.723 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 40/4 = 10 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (10, 6) = 10 mm
               Provide 10 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 40 = 640 mm
               iii) 300 mm
               Required spacing = minimum of (300, 640, 300) = 300 mm
# Provide ties               10 @ 300 mm c/c

SUMMARY :

CG22 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #40 - 8 nos. (10053 mm˛)
Provide ties #10 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG23 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 564.78 kN
MomentX,(Mx) = 21.59 kN-m
MomentY,(My) = 8.55 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C310.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 11.691 kN.m
My_MinEccen = 11.296 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 21.594 kN.m
My = max(My,My_MinEccen) + MuaddY = 11.296 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 21.59/564.78 = 38 mm
Actual eccenY = My / P = 8.55/564.78 = 15 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(38,21) = 38 mm
eccenY = max(Actual eccenY,eccenYMin) = max(15,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(38 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00285 352.55 8.92 343.63 77.73 179 13.91
2 226 0.00157 301.29 8.51 292.78 66.22 90 5.93
3 226 0.00030 60.00 2.48 57.52 13.01 0 0.00
4 226 -0.00097 -194.58 0.00 -194.58 -44.01 -90 3.94
5 226 -0.00225 -336.77 0.00 -336.77 -76.17 -179 13.64
Total36.78 37.41
xux = 246 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 246/450 = 0.197
Puc = 0.197 x 20.00 x 300 x 450
= 531.56 kN
Pux1 = Puc + Pus(Total)
= 531.56 + (36.78)
= 568.34 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 246/450
= 0.228
Muc = 531.56 x (0.5 x 450 - 0.228 x 450) = 65.18 kN-m
Mux1 = Muc + Mus(Total)
= 65.18 + (37.41)
= 102.60 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00256 346.63 8.92 337.71 190.97 104 19.86
2 565 -0.00170 -310.82 0.00 -310.82 -175.76 -104 18.28
Total15.21 38.14
xuy = 171 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 171/300 = 0.205
Puc = 0.205 x 20.00 x 450 x 300
= 554.34 kN
Puy1 = Puc + Pus(Total)
= 554.34 + (15.21)
= 569.55 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 171/300
= 0.237
Muc = 554.34 x (0.5 x 300 - 0.237 x 300) = 43.70 kN-m
Muy1 = Muc + Mus(Total)
= 43.70 + (38.14)
= 81.84 kN-m
Pu/Puz = 0.363
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.363, an = 1.271
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((21.59/102.60)1.271) + ((11.30/81.84)1.271)
= 0.219 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG23 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1103.16 kN
MomentX,(Mx) = 22.48 kN-m
MomentY,(My) = 8.68 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C310.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 22.835 kN.m
My_MinEccen = 22.063 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 22.835 kN.m
My = max(My,My_MinEccen) + MuaddY = 22.063 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 22.48/1103.16 = 20 mm
Actual eccenY = My / P = 8.68/1103.16 = 8 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(20,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(8,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00309 354.73 8.92 345.81 78.22 179 14.00
2 226 0.00231 338.99 8.92 330.07 74.66 90 6.68
3 226 0.00152 296.06 8.40 287.66 65.07 0 0.00
4 226 0.00073 145.84 5.32 140.52 31.78 -90 -2.84
5 226 -0.00006 -11.87 0.00 -11.87 -2.68 -179 0.48
Total247.05 18.32
xux = 397 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 397/450 = 0.318
Puc = 0.318 x 20.00 x 300 x 450
= 858.09 kN
Pux1 = Puc + Pus(Total)
= 858.09 + (247.05)
= 1105.14 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 397/450
= 0.367
Muc = 858.09 x (0.5 x 450 - 0.367 x 450) = 51.26 kN-m
Mux1 = Muc + Mus(Total)
= 51.26 + (18.32)
= 69.58 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00291 353.14 8.92 344.22 194.65 104 20.24
2 565 0.00026 51.61 2.15 49.46 27.97 -104 -2.91
Total222.62 17.33
xuy = 274 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 274/300 = 0.329
Puc = 0.329 x 20.00 x 450 x 300
= 888.47 kN
Puy1 = Puc + Pus(Total)
= 888.47 + (222.62)
= 1111.09 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 274/300
= 0.380
Muc = 888.47 x (0.5 x 300 - 0.380 x 300) = 31.92 kN-m
Muy1 = Muc + Mus(Total)
= 31.92 + (17.33)
= 49.25 kN-m
Pu/Puz = 0.709
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.709, an = 1.848
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((22.84/69.58)1.848) + ((22.06/49.25)1.848)
= 0.354 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG23 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1664.34 kN
MomentX,(Mx) = 21.03 kN-m
MomentY,(My) = 9.60 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C310.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 34.452 kN.m
My_MinEccen = 33.287 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 34.452 kN.m
My = max(My,My_MinEccen) + MuaddY = 33.287 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 21.03/1664.34 = 13 mm
Actual eccenY = My / P = 9.60/1664.34 = 6 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(13,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(6,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 14 nos. (2815 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00312 354.96 8.92 346.04 139.15 177 24.63
2 402 0.00265 349.09 8.92 340.17 136.79 118 16.14
3 402 0.00219 334.66 8.92 325.74 130.99 59 7.73
4 402 0.00172 312.46 8.75 303.71 122.13 0 0.00
5 402 0.00126 251.21 7.69 243.52 97.93 -59 -5.78
6 402 0.00079 157.98 5.65 152.32 61.25 -118 -7.23
7 402 0.00032 64.74 2.65 62.09 24.97 -177 -4.42
Total713.20 31.07
xux = 443 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 443/450 = 0.354
Puc = 0.354 x 20.00 x 300 x 450
= 956.81 kN
Pux1 = Puc + Pus(Total)
= 956.81 + (713.20)
= 1670.01 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 443/450
= 0.410
Muc = 956.81 x (0.5 x 450 - 0.410 x 450) = 38.97 kN-m
Mux1 = Muc + Mus(Total)
= 38.97 + (31.07)
= 70.04 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1407 0.00287 352.75 8.92 343.83 483.92 102 49.36
2 1407 0.00067 133.83 4.97 128.86 181.37 -102 -18.50
Total665.29 30.86
xuy = 314 mm               Puc = C1.fck.B.D
ku = 1.047
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.976
C1 = 0.446 x (1 - C3/6) = 0.373
Puc = 0.373 x 20.00 x 450 x 300
= 1008.29 kN
Puy1 = Puc + Pus(Total)
= 1008.29 + (665.29)
= 1673.58 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.431
Muc = 1008.29 x (0.5 x 300 - 0.431 x 300) = 20.99 kN-m
Muy1 = Muc + Mus(Total)
= 20.99 + (30.86)
= 51.85 kN-m
Pu/Puz = 0.806
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.806, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((34.45/70.04)2.000) + ((33.29/51.85)2.000)
= 0.654 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG23 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 14 nos. (2815 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 2214.86 kN
MomentX,(Mx) = 16.45 kN-m
MomentY,(My) = 10.91 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C310.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 45.848 kN.m
My_MinEccen = 44.297 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 45.848 kN.m
My = max(My,My_MinEccen) + MuaddY = 44.297 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 16.45/2214.86 = 7 mm
Actual eccenY = My / P = 10.91/2214.86 = 5 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(7,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(5,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 16 nos. (5027 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00310 354.79 8.92 345.87 217.32 175 38.03
2 628 0.00270 350.35 8.92 341.43 214.53 125 26.82
3 628 0.00231 338.96 8.92 330.04 207.37 75 15.55
4 628 0.00191 324.00 8.90 315.09 197.98 25 4.95
5 628 0.00151 295.23 8.38 286.84 180.23 -25 -4.51
6 628 0.00111 222.13 7.16 214.98 135.07 -75 -10.13
7 628 0.00071 142.49 5.22 137.27 86.25 -125 -10.78
8 628 0.00031 62.84 2.58 60.26 37.86 -175 -6.63
Total1276.60 53.30
xux = 439 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 439/450 = 0.352
Puc = 0.352 x 20.00 x 300 x 450
= 949.22 kN
Pux1 = Puc + Pus(Total)
= 949.22 + (1276.60)
= 2225.82 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 439/450
= 0.406
Muc = 949.22 x (0.5 x 450 - 0.406 x 450) = 40.05 kN-m
Mux1 = Muc + Mus(Total)
= 40.05 + (53.30)
= 93.35 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 2513 0.00283 352.38 8.92 343.46 863.21 100 86.32
2 2513 0.00072 144.60 5.28 139.32 350.14 -100 -35.01
Total1213.35 51.31
xuy = 319 mm               Puc = C1.fck.B.D
ku = 1.063
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.929
C1 = 0.446 x (1 - C3/6) = 0.377
Puc = 0.377 x 20.00 x 450 x 300
= 1017.83 kN
Puy1 = Puc + Pus(Total)
= 1017.83 + (1213.35)
= 2231.18 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.435
Muc = 1017.83 x (0.5 x 300 - 0.435 x 300) = 19.97 kN-m
Muy1 = Muc + Mus(Total)
= 19.97 + (51.31)
= 71.28 kN-m
Pu/Puz = 0.810
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.810, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((45.85/93.35)2.000) + ((44.30/71.28)2.000)
= 0.627 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 20 = 320 mm
               iii) 300 mm
               Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG23 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 16 nos. (5027 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 2856.10 kN
MomentX,(Mx) = 8.71 kN-m
MomentY,(My) = 7.68 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C310.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 62.120 kN.m
My_MinEccen = 57.122 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 62.120 kN.m
My = max(My,My_MinEccen) + MuaddY = 57.122 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 8.71/2856.10 = 3 mm
Actual eccenY = My / P = 7.68/2856.10 = 3 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(3,22) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(3,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #32 - 10 nos. (8042 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1608 0.00304 354.24 8.92 345.32 555.45 169 93.87
2 1608 0.00234 340.38 8.92 331.46 533.14 85 45.05
3 1608 0.00165 307.87 8.64 299.23 481.30 0 0.00
4 1608 0.00095 190.70 6.48 184.23 296.33 -85 -25.04
5 1608 0.00026 51.65 2.16 49.50 79.62 -169 -13.46
Total1945.85 100.43
xux = 425 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 425/450 = 0.340
Puc = 0.340 x 20.00 x 300 x 450
= 918.84 kN
Pux1 = Puc + Pus(Total)
= 918.84 + (1945.85)
= 2864.69 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 425/450
= 0.393
Muc = 918.84 x (0.5 x 450 - 0.393 x 450) = 44.14 kN-m
Mux1 = Muc + Mus(Total)
= 44.14 + (100.43)
= 144.57 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 4021 0.00285 352.56 8.92 343.64 1381.85 94 129.89
2 4021 0.00065 130.67 4.88 125.79 505.83 -94 -47.55
Total1887.69 82.35
xuy = 300 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 300/300 = 0.360
Puc = 0.360 x 20.00 x 450 x 300
= 972.00 kN
Puy1 = Puc + Pus(Total)
= 972.00 + (1887.69)
= 2859.69 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 300/300
= 0.416
Muc = 972.00 x (0.5 x 300 - 0.416 x 300) = 24.49 kN-m
Muy1 = Muc + Mus(Total)
= 24.49 + (82.35)
= 106.84 kN-m
Pu/Puz = 0.783
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.783, an = 1.972
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((62.12/144.57)1.972) + ((57.12/106.84)1.972)
= 0.480 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 32/4 = 8 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (8, 6) = 8 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 32 = 512 mm
               iii) 300 mm
               Required spacing = minimum of (300, 512, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG23 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #32 - 10 nos. (8042 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG24 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 356.17 kN
MomentX,(Mx) = 5.25 kN-m
MomentY,(My) = 74.25 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C320.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 7.373 kN.m
My_MinEccen = 7.123 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 7.373 kN.m
My = max(My,My_MinEccen) + MuaddY = 74.253 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 5.25/356.17 = 15 mm
Actual eccenY = My / P = 74.25/356.17 = 208 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(15,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(208,20) = 208 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(208 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 6 nos. (1206 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00263 348.38 8.92 339.46 136.51 177 24.16
2 402 -0.00059 -118.26 0.00 -118.26 -47.56 0 0.00
3 402 -0.00381 -360.90 0.00 -360.90 -145.13 -177 25.69
Total-56.18 49.85
xux = 192 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 192/450 = 0.154
Puc = 0.154 x 20.00 x 300 x 450
= 415.76 kN
Pux1 = Puc + Pus(Total)
= 415.76 + (-56.18)
= 359.58 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 192/450
= 0.178
Muc = 415.76 x (0.5 x 450 - 0.178 x 450) = 60.26 kN-m
Mux1 = Muc + Mus(Total)
= 60.26 + (49.85)
= 110.10 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 603 0.00207 330.44 8.92 321.52 193.94 102 19.78
2 603 -0.00399 -360.90 0.00 -360.90 -217.69 -102 22.20
Total-23.75 41.99
xuy = 118 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 118/300 = 0.141
Puc = 0.141 x 20.00 x 450 x 300
= 381.59 kN
Puy1 = Puc + Pus(Total)
= 381.59 + (-23.75)
= 357.83 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 118/300
= 0.163
Muc = 381.59 x (0.5 x 300 - 0.163 x 300) = 38.54 kN-m
Muy1 = Muc + Mus(Total)
= 38.54 + (41.99)
= 80.53 kN-m
Pu/Puz = 0.225
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.225, an = 1.042
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((7.37/110.10)1.042) + ((74.25/80.53)1.042)
= 0.979 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG24 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 6 nos. (1206 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 732.77 kN
MomentX,(Mx) = 4.75 kN-m
MomentY,(My) = 67.56 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C320.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 15.168 kN.m
My_MinEccen = 14.655 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 15.168 kN.m
My = max(My,My_MinEccen) + MuaddY = 67.556 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 4.75/732.77 = 6 mm
Actual eccenY = My / P = 67.56/732.77 = 92 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(6,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(92,20) = 92 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(92 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00294 353.42 8.92 344.50 77.92 179 13.95
2 226 0.00186 321.36 8.88 312.48 70.68 90 6.33
3 226 0.00078 156.97 5.63 151.34 34.23 0 0.00
4 226 -0.00030 -59.04 0.00 -59.04 -13.35 -90 1.20
5 226 -0.00138 -275.04 0.00 -275.04 -62.21 -179 11.14
Total107.27 32.61
xux = 290 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 290/450 = 0.232
Puc = 0.232 x 20.00 x 300 x 450
= 626.48 kN
Pux1 = Puc + Pus(Total)
= 626.48 + (107.27)
= 733.76 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 290/450
= 0.268
Muc = 626.48 x (0.5 x 450 - 0.268 x 450) = 65.37 kN-m
Mux1 = Muc + Mus(Total)
= 65.37 + (32.61)
= 97.98 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00270 350.17 8.92 341.25 192.97 104 20.07
2 565 -0.00094 -187.27 0.00 -187.27 -105.90 -104 11.01
Total87.07 31.08
xuy = 200 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 200/300 = 0.240
Puc = 0.240 x 20.00 x 450 x 300
= 649.27 kN
Puy1 = Puc + Pus(Total)
= 649.27 + (87.07)
= 736.34 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 200/300
= 0.278
Muc = 649.27 x (0.5 x 300 - 0.278 x 300) = 43.27 kN-m
Muy1 = Muc + Mus(Total)
= 43.27 + (31.08)
= 74.35 kN-m
Pu/Puz = 0.471
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.471, an = 1.451
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((15.17/97.98)1.451) + ((67.56/74.35)1.451)
= 0.937 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG24 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1102.81 kN
MomentX,(Mx) = 4.76 kN-m
MomentY,(My) = 64.78 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C320.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 22.828 kN.m
My_MinEccen = 22.056 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 22.828 kN.m
My = max(My,My_MinEccen) + MuaddY = 64.779 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 4.76/1102.81 = 4 mm
Actual eccenY = My / P = 64.78/1102.81 = 59 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(4,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(59,20) = 59 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(59 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 16 nos. (1810 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00305 354.31 8.92 345.39 78.12 179 13.98
2 226 0.00254 346.21 8.92 337.29 76.29 128 9.75
3 226 0.00204 329.15 8.92 320.23 72.43 77 5.56
4 226 0.00153 297.63 8.44 289.19 65.41 26 1.67
5 226 0.00103 206.02 6.82 199.20 45.06 -26 -1.15
6 226 0.00053 105.20 4.07 101.13 22.87 -77 -1.75
7 226 0.00002 4.38 0.19 4.18 0.95 -128 -0.12
8 226 -0.00048 -96.44 0.00 -96.44 -21.82 -179 3.90
Total339.33 31.85
xux = 355 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 355/450 = 0.284
Puc = 0.284 x 20.00 x 300 x 450
= 766.97 kN
Pux1 = Puc + Pus(Total)
= 766.97 + (339.33)
= 1106.30 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 355/450
= 0.328
Muc = 766.97 x (0.5 x 450 - 0.328 x 450) = 59.28 kN-m
Mux1 = Muc + Mus(Total)
= 59.28 + (31.85)
= 91.12 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 905 0.00285 352.63 8.92 343.71 310.98 104 32.34
2 905 -0.00006 -12.31 0.00 -12.31 -11.14 -104 1.16
Total299.84 33.50
xuy = 250 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 250/300 = 0.300
Puc = 0.300 x 20.00 x 450 x 300
= 808.73 kN
Puy1 = Puc + Pus(Total)
= 808.73 + (299.84)
= 1108.58 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 250/300
= 0.346
Muc = 808.73 x (0.5 x 300 - 0.346 x 300) = 37.33 kN-m
Muy1 = Muc + Mus(Total)
= 37.33 + (33.50)
= 70.83 kN-m
Pu/Puz = 0.626
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.626, an = 1.710
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((22.83/91.12)1.710) + ((64.78/70.83)1.710)
= 0.952 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG24 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 16 nos. (1810 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1463.17 kN
MomentX,(Mx) = 4.88 kN-m
MomentY,(My) = 60.78 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C320.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 30.288 kN.m
My_MinEccen = 29.263 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 30.288 kN.m
My = max(My,My_MinEccen) + MuaddY = 60.780 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 4.88/1463.17 = 3 mm
Actual eccenY = My / P = 60.78/1463.17 = 42 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(3,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(42,20) = 42 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(42 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 14 nos. (2815 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00307 354.54 8.92 345.62 138.98 177 24.60
2 402 0.00255 346.37 8.92 337.45 135.70 118 16.01
3 402 0.00202 328.64 8.92 319.72 128.57 59 7.59
4 402 0.00150 294.38 8.36 286.02 115.02 0 0.00
5 402 0.00098 195.11 6.58 188.53 75.81 -59 -4.47
6 402 0.00045 90.22 3.57 86.65 34.84 -118 -4.11
7 402 -0.00007 -14.67 0.00 -14.67 -5.90 -177 1.04
Total623.02 40.66
xux = 394 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 394/450 = 0.315
Puc = 0.315 x 20.00 x 300 x 450
= 850.50 kN
Pux1 = Puc + Pus(Total)
= 850.50 + (623.02)
= 1473.52 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 394/450
= 0.364
Muc = 850.50 x (0.5 x 450 - 0.364 x 450) = 52.05 kN-m
Mux1 = Muc + Mus(Total)
= 52.05 + (40.66)
= 92.71 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1407 0.00289 352.96 8.92 344.04 484.21 102 49.39
2 1407 0.00031 62.17 2.56 59.61 83.90 -102 -8.56
Total568.11 40.83
xuy = 277 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 277/300 = 0.332
Puc = 0.332 x 20.00 x 450 x 300
= 896.06 kN
Puy1 = Puc + Pus(Total)
= 896.06 + (568.11)
= 1464.18 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 277/300
= 0.384
Muc = 896.06 x (0.5 x 300 - 0.384 x 300) = 31.32 kN-m
Muy1 = Muc + Mus(Total)
= 31.32 + (40.83)
= 72.15 kN-m
Pu/Puz = 0.708
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.708, an = 1.847
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((30.29/92.71)1.847) + ((60.78/72.15)1.847)
= 0.855 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG24 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 14 nos. (2815 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 1807.30 kN
MomentX,(Mx) = 2.66 kN-m
MomentY,(My) = 39.45 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C320.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 39.309 kN.m
My_MinEccen = 36.146 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 39.309 kN.m
My = max(My,My_MinEccen) + MuaddY = 39.454 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 2.66/1807.30 = 1 mm
Actual eccenY = My / P = 39.45/1807.30 = 22 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(1,22) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(22,20) = 22 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(22 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 16 nos. (3217 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00311 354.88 8.92 345.96 139.12 177 24.62
2 402 0.00272 350.86 8.92 341.94 137.50 126 17.38
3 402 0.00234 340.06 8.92 331.14 133.16 76 10.10
4 402 0.00195 325.81 8.91 316.89 127.43 25 3.22
5 402 0.00156 300.01 8.49 291.52 117.23 -25 -2.96
6 402 0.00117 234.27 7.39 226.88 91.23 -76 -6.92
7 402 0.00078 156.66 5.62 151.04 60.74 -126 -7.68
8 402 0.00040 79.05 3.18 75.88 30.51 -177 -5.40
Total836.92 32.37
xux = 454 mm               Puc = C1.fck.B.D
ku = 1.008
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.112
C1 = 0.446 x (1 - C3/6) = 0.363
Puc = 0.363 x 20.00 x 300 x 450
= 980.97 kN
Pux1 = Puc + Pus(Total)
= 980.97 + (836.92)
= 1817.89 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.419
Muc = 980.97 x (0.5 x 450 - 0.419 x 450) = 35.88 kN-m
Mux1 = Muc + Mus(Total)
= 35.88 + (32.37)
= 68.24 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1608 0.00281 352.22 8.92 343.30 552.19 102 56.32
2 1608 0.00076 152.59 5.51 147.08 236.58 -102 -24.13
Total788.77 32.19
xuy = 328 mm               Puc = C1.fck.B.D
ku = 1.094
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.843
C1 = 0.446 x (1 - C3/6) = 0.383
Puc = 0.383 x 20.00 x 450 x 300
= 1034.93 kN
Puy1 = Puc + Pus(Total)
= 1034.93 + (788.77)
= 1823.70 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.442
Muc = 1034.93 x (0.5 x 300 - 0.442 x 300) = 18.14 kN-m
Muy1 = Muc + Mus(Total)
= 18.14 + (32.19)
= 50.33 kN-m
Pu/Puz = 0.826
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.826, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((39.31/68.24)2.000) + ((39.45/50.33)2.000)
= 0.946 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG24 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 16 nos. (3217 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG25 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 373.62 kN
MomentX,(Mx) = 9.78 kN-m
MomentY,(My) = 86.66 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C330.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 7.734 kN.m
My_MinEccen = 7.472 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 9.779 kN.m
My = max(My,My_MinEccen) + MuaddY = 86.655 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 9.78/373.62 = 26 mm
Actual eccenY = My / P = 86.66/373.62 = 232 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(26,21) = 26 mm
eccenY = max(Actual eccenY,eccenYMin) = max(232,20) = 232 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(26 mm) > 0.05 x 300(15 mm)
and eccenY(232 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 14 nos. (1583 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00271 350.53 8.92 341.61 77.27 179 13.83
2 226 0.00169 310.21 8.70 301.51 68.20 119 8.14
3 226 0.00066 132.42 4.93 127.49 28.84 60 1.72
4 226 -0.00036 -72.41 0.00 -72.41 -16.38 -0 0.00
5 226 -0.00139 -277.25 0.00 -277.25 -62.71 -60 3.74
6 226 -0.00241 -342.81 0.00 -342.81 -77.54 -119 9.25
7 226 -0.00343 -357.70 0.00 -357.70 -80.91 -179 14.48
Total-63.24 51.17
xux = 204 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 204/450 = 0.163
Puc = 0.163 x 20.00 x 300 x 450
= 440.44 kN
Pux1 = Puc + Pus(Total)
= 440.44 + (-63.24)
= 377.20 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 204/450
= 0.189
Muc = 440.44 x (0.5 x 450 - 0.189 x 450) = 61.74 kN-m
Mux1 = Muc + Mus(Total)
= 61.74 + (51.17)
= 112.91 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 792 0.00220 335.23 8.92 326.31 258.33 104 26.87
2 792 -0.00366 -359.65 0.00 -359.65 -284.73 -104 29.61
Total-26.39 56.48
xuy = 124 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 124/300 = 0.149
Puc = 0.149 x 20.00 x 450 x 300
= 402.47 kN
Puy1 = Puc + Pus(Total)
= 402.47 + (-26.39)
= 376.08 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 124/300
= 0.172
Muc = 402.47 x (0.5 x 300 - 0.172 x 300) = 39.57 kN-m
Muy1 = Muc + Mus(Total)
= 39.57 + (56.48)
= 96.05 kN-m
Pu/Puz = 0.221
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.221, an = 1.034
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((9.78/112.91)1.034) + ((86.66/96.05)1.034)
= 0.979 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG25 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 14 nos. (1583 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 768.10 kN
MomentX,(Mx) = 8.18 kN-m
MomentY,(My) = 79.01 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C330.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 15.900 kN.m
My_MinEccen = 15.362 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 15.900 kN.m
My = max(My,My_MinEccen) + MuaddY = 79.006 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 8.18/768.10 = 11 mm
Actual eccenY = My / P = 79.01/768.10 = 103 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(11,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(103,20) = 103 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(103 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 14 nos. (1583 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00294 353.36 8.92 344.44 77.91 179 13.95
2 226 0.00221 335.43 8.92 326.51 73.85 119 8.81
3 226 0.00148 292.53 8.32 284.21 64.29 60 3.84
4 226 0.00075 150.31 5.44 144.86 32.77 -0 -0.00
5 226 0.00002 4.54 0.20 4.34 0.98 -60 -0.06
6 226 -0.00071 -141.23 0.00 -141.23 -31.95 -119 3.81
7 226 -0.00144 -287.00 0.00 -287.00 -64.92 -179 11.62
Total152.93 41.97
xux = 287 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 287/450 = 0.229
Puc = 0.229 x 20.00 x 300 x 450
= 618.89 kN
Pux1 = Puc + Pus(Total)
= 618.89 + (152.93)
= 771.82 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 287/450
= 0.265
Muc = 618.89 x (0.5 x 450 - 0.265 x 450) = 65.48 kN-m
Mux1 = Muc + Mus(Total)
= 65.48 + (41.97)
= 107.45 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 792 0.00270 350.17 8.92 341.25 270.16 104 28.10
2 792 -0.00094 -187.27 0.00 -187.27 -148.26 -104 15.42
Total121.90 43.52
xuy = 200 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 200/300 = 0.240
Puc = 0.240 x 20.00 x 450 x 300
= 649.27 kN
Puy1 = Puc + Pus(Total)
= 649.27 + (121.90)
= 771.17 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 200/300
= 0.278
Muc = 649.27 x (0.5 x 300 - 0.278 x 300) = 43.27 kN-m
Muy1 = Muc + Mus(Total)
= 43.27 + (43.52)
= 86.78 kN-m
Pu/Puz = 0.454
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.454, an = 1.423
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((15.90/107.45)1.423) + ((79.01/86.78)1.423)
= 0.941 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG25 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 14 nos. (1583 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1162.53 kN
MomentX,(Mx) = 6.58 kN-m
MomentY,(My) = 74.59 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C330.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 24.064 kN.m
My_MinEccen = 23.251 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 24.064 kN.m
My = max(My,My_MinEccen) + MuaddY = 74.586 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 6.58/1162.53 = 6 mm
Actual eccenY = My / P = 74.59/1162.53 = 64 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(6,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(64,20) = 64 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(64 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 12 nos. (2413 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00301 354.01 8.92 345.09 138.77 177 24.56
2 402 0.00229 338.51 8.92 329.59 132.53 106 14.08
3 402 0.00157 301.39 8.52 292.87 117.77 35 4.17
4 402 0.00085 170.93 5.99 164.94 66.33 -35 -2.35
5 402 0.00014 27.09 1.17 25.92 10.42 -106 -1.11
6 402 -0.00058 -116.76 0.00 -116.76 -46.95 -177 8.31
Total418.87 47.66
xux = 345 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 345/450 = 0.276
Puc = 0.276 x 20.00 x 300 x 450
= 744.19 kN
Pux1 = Puc + Pus(Total)
= 744.19 + (418.87)
= 1163.06 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 345/450
= 0.319
Muc = 744.19 x (0.5 x 450 - 0.319 x 450) = 60.78 kN-m
Mux1 = Muc + Mus(Total)
= 60.78 + (47.66)
= 108.44 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1206 0.00281 352.22 8.92 343.30 414.14 102 42.24
2 1206 -0.00014 -27.19 0.00 -27.19 -32.80 -102 3.35
Total381.34 45.59
xuy = 243 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 243/300 = 0.291
Puc = 0.291 x 20.00 x 450 x 300
= 785.95 kN
Puy1 = Puc + Pus(Total)
= 785.95 + (381.34)
= 1167.30 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 243/300
= 0.336
Muc = 785.95 x (0.5 x 300 - 0.336 x 300) = 38.58 kN-m
Muy1 = Muc + Mus(Total)
= 38.58 + (45.59)
= 84.17 kN-m
Pu/Puz = 0.598
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.598, an = 1.663
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((24.06/108.44)1.663) + ((74.59/84.17)1.663)
= 0.900 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG25 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 12 nos. (2413 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1557.77 kN
MomentX,(Mx) = 4.33 kN-m
MomentY,(My) = 68.24 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C330.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 32.246 kN.m
My_MinEccen = 31.155 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 32.246 kN.m
My = max(My,My_MinEccen) + MuaddY = 68.240 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 4.33/1557.77 = 3 mm
Actual eccenY = My / P = 68.24/1557.77 = 44 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(3,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(44,20) = 44 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(44 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 16 nos. (3217 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00307 354.54 8.92 345.62 138.98 177 24.60
2 402 0.00262 348.30 8.92 339.38 136.47 126 17.25
3 402 0.00217 334.14 8.92 325.22 130.78 76 9.92
4 402 0.00172 312.61 8.75 303.86 122.19 25 3.09
5 402 0.00128 255.05 7.75 247.30 99.44 -25 -2.51
6 402 0.00083 165.14 5.84 159.30 64.06 -76 -4.86
7 402 0.00038 75.24 3.04 72.20 29.03 -126 -3.67
8 402 -0.00007 -14.67 0.00 -14.67 -5.90 -177 1.04
Total715.06 44.86
xux = 394 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 394/450 = 0.315
Puc = 0.315 x 20.00 x 300 x 450
= 850.50 kN
Pux1 = Puc + Pus(Total)
= 850.50 + (715.06)
= 1565.56 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 394/450
= 0.364
Muc = 850.50 x (0.5 x 450 - 0.364 x 450) = 52.05 kN-m
Mux1 = Muc + Mus(Total)
= 52.05 + (44.86)
= 96.91 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1608 0.00290 353.00 8.92 344.08 553.46 102 56.45
2 1608 0.00034 67.53 2.76 64.77 104.19 -102 -10.63
Total657.64 45.83
xuy = 279 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 279/300 = 0.335
Puc = 0.335 x 20.00 x 450 x 300
= 903.66 kN
Puy1 = Puc + Pus(Total)
= 903.66 + (657.64)
= 1561.30 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 279/300
= 0.387
Muc = 903.66 x (0.5 x 300 - 0.387 x 300) = 30.70 kN-m
Muy1 = Muc + Mus(Total)
= 30.70 + (45.83)
= 76.53 kN-m
Pu/Puz = 0.712
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.712, an = 1.854
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((32.25/96.91)1.854) + ((68.24/76.53)1.854)
= 0.939 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG25 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 16 nos. (3217 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 1952.82 kN
MomentX,(Mx) = 1.88 kN-m
MomentY,(My) = 43.19 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C330.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 42.474 kN.m
My_MinEccen = 39.056 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 42.474 kN.m
My = max(My,My_MinEccen) + MuaddY = 43.191 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 1.88/1952.82 = 1 mm
Actual eccenY = My / P = 43.19/1952.82 = 22 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(1,22) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(22,20) = 22 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(22 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 12 nos. (3770 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00308 354.61 8.92 345.69 217.21 175 38.01
2 628 0.00255 346.44 8.92 337.52 212.07 105 22.27
3 628 0.00202 328.53 8.92 319.61 200.82 35 7.03
4 628 0.00149 293.60 8.34 285.25 179.23 -35 -6.27
5 628 0.00096 192.34 6.52 185.83 116.76 -105 -12.26
6 628 0.00043 86.35 3.44 82.92 52.10 -175 -9.12
Total978.18 39.66
xux = 457 mm               Puc = C1.fck.B.D
ku = 1.016
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.083
C1 = 0.446 x (1 - C3/6) = 0.366
Puc = 0.366 x 20.00 x 300 x 450
= 986.88 kN
Pux1 = Puc + Pus(Total)
= 986.88 + (978.18)
= 1965.06 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.421
Muc = 986.88 x (0.5 x 450 - 0.421 x 450) = 34.93 kN-m
Mux1 = Muc + Mus(Total)
= 34.93 + (39.66)
= 74.58 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1885 0.00279 352.04 8.92 343.12 646.77 100 64.68
2 1885 0.00078 156.60 5.62 150.98 284.60 -100 -28.46
Total931.36 36.22
xuy = 328 mm               Puc = C1.fck.B.D
ku = 1.094
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.843
C1 = 0.446 x (1 - C3/6) = 0.383
Puc = 0.383 x 20.00 x 450 x 300
= 1034.93 kN
Puy1 = Puc + Pus(Total)
= 1034.93 + (931.36)
= 1966.29 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.442
Muc = 1034.93 x (0.5 x 300 - 0.442 x 300) = 18.14 kN-m
Muy1 = Muc + Mus(Total)
= 18.14 + (36.22)
= 54.35 kN-m
Pu/Puz = 0.829
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.829, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((42.47/74.58)2.000) + ((43.19/54.35)2.000)
= 0.956 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 20 = 320 mm
               iii) 300 mm
               Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG25 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 12 nos. (3770 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG26 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 793.09 kN
MomentX,(Mx) = 24.78 kN-m
MomentY,(My) = 11.05 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C340.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 16.417 kN.m
My_MinEccen = 15.862 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 24.775 kN.m
My = max(My,My_MinEccen) + MuaddY = 15.862 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 24.78/793.09 = 31 mm
Actual eccenY = My / P = 11.05/793.09 = 14 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(31,21) = 31 mm
eccenY = max(Actual eccenY,eccenYMin) = max(14,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(31 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00297 353.67 8.92 344.75 77.98 179 13.96
2 226 0.00195 325.88 8.91 316.97 71.70 90 6.42
3 226 0.00093 185.06 6.34 178.71 40.42 0 0.00
4 226 -0.00010 -19.78 0.00 -19.78 -4.47 -90 0.40
5 226 -0.00112 -224.61 0.00 -224.61 -50.81 -179 9.09
Total134.82 29.87
xux = 306 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 306/450 = 0.245
Puc = 0.245 x 20.00 x 300 x 450
= 660.66 kN
Pux1 = Puc + Pus(Total)
= 660.66 + (134.82)
= 795.48 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 306/450
= 0.283
Muc = 660.66 x (0.5 x 450 - 0.283 x 450) = 64.59 kN-m
Mux1 = Muc + Mus(Total)
= 64.59 + (29.87)
= 94.46 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00274 351.20 8.92 342.28 193.56 104 20.13
2 565 -0.00071 -142.90 0.00 -142.90 -80.81 -104 8.40
Total112.75 28.53
xuy = 211 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 211/300 = 0.253
Puc = 0.253 x 20.00 x 450 x 300
= 683.44 kN
Puy1 = Puc + Pus(Total)
= 683.44 + (112.75)
= 796.18 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 211/300
= 0.293
Muc = 683.44 x (0.5 x 300 - 0.293 x 300) = 42.54 kN-m
Muy1 = Muc + Mus(Total)
= 42.54 + (28.53)
= 71.08 kN-m
Pu/Puz = 0.509
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.509, an = 1.516
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((24.78/94.46)1.516) + ((15.86/71.08)1.516)
= 0.235 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG26 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1575.90 kN
MomentX,(Mx) = 20.28 kN-m
MomentY,(My) = 11.14 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C340.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 32.621 kN.m
My_MinEccen = 31.518 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 32.621 kN.m
My = max(My,My_MinEccen) + MuaddY = 31.518 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 20.28/1575.90 = 13 mm
Actual eccenY = My / P = 11.14/1575.90 = 7 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(13,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(7,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 20 nos. (2262 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00311 354.88 8.92 345.96 78.25 179 14.01
2 226 0.00281 352.24 8.92 343.32 77.66 139 10.81
3 226 0.00251 345.36 8.92 336.44 76.10 99 7.57
4 226 0.00221 335.39 8.92 326.47 73.85 60 4.41
5 226 0.00191 324.00 8.90 315.10 71.27 20 1.42
6 226 0.00161 304.43 8.57 295.86 66.92 -20 -1.33
7 226 0.00130 260.99 7.84 253.14 57.26 -60 -3.42
8 226 0.00100 200.76 6.71 194.05 43.89 -99 -4.36
9 226 0.00070 140.53 5.17 135.36 30.62 -139 -4.26
10 226 0.00040 80.30 3.22 77.08 17.43 -179 -3.12
Total593.26 21.71
xux = 457 mm               Puc = C1.fck.B.D
ku = 1.016
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.083
C1 = 0.446 x (1 - C3/6) = 0.366
Puc = 0.366 x 20.00 x 300 x 450
= 986.88 kN
Pux1 = Puc + Pus(Total)
= 986.88 + (593.26)
= 1580.14 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.421
Muc = 986.88 x (0.5 x 450 - 0.421 x 450) = 34.93 kN-m
Mux1 = Muc + Mus(Total)
= 34.93 + (21.71)
= 56.64 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1131 0.00283 352.39 8.92 343.47 388.46 104 40.40
2 1131 0.00074 148.58 5.40 143.19 161.94 -104 -16.84
Total550.40 23.56
xuy = 328 mm               Puc = C1.fck.B.D
ku = 1.094
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.843
C1 = 0.446 x (1 - C3/6) = 0.383
Puc = 0.383 x 20.00 x 450 x 300
= 1034.93 kN
Puy1 = Puc + Pus(Total)
= 1034.93 + (550.40)
= 1585.32 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.442
Muc = 1034.93 x (0.5 x 300 - 0.442 x 300) = 18.14 kN-m
Muy1 = Muc + Mus(Total)
= 18.14 + (23.56)
= 41.69 kN-m
Pu/Puz = 0.830
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.830, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((32.62/56.64)2.000) + ((31.52/41.69)2.000)
= 0.903 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG26 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 20 nos. (2262 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 2390.14 kN
MomentX,(Mx) = 15.68 kN-m
MomentY,(My) = 10.75 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C340.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 49.476 kN.m
My_MinEccen = 47.803 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 49.476 kN.m
My = max(My,My_MinEccen) + MuaddY = 47.803 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 15.68/2390.14 = 7 mm
Actual eccenY = My / P = 10.75/2390.14 = 4 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(7,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(4,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #25 - 12 nos. (5890 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 982 0.00308 354.56 8.92 345.64 339.33 173 58.53
2 982 0.00252 345.54 8.92 336.62 330.48 104 34.20
3 982 0.00196 326.20 8.92 317.28 311.49 35 10.75
4 982 0.00140 279.92 8.12 271.81 266.85 -35 -9.21
5 982 0.00084 168.23 5.93 162.30 159.34 -104 -16.49
6 982 0.00028 56.53 2.34 54.19 53.20 -173 -9.18
Total1460.68 68.61
xux = 432 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 432/450 = 0.346
Puc = 0.346 x 20.00 x 300 x 450
= 934.03 kN
Pux1 = Puc + Pus(Total)
= 934.03 + (1460.68)
= 2394.71 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 432/450
= 0.400
Muc = 934.03 x (0.5 x 450 - 0.400 x 450) = 42.14 kN-m
Mux1 = Muc + Mus(Total)
= 42.14 + (68.61)
= 110.75 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 2945 0.00284 352.51 8.92 343.59 1011.96 98 98.67
2 2945 0.00068 136.89 5.06 131.83 388.27 -98 -37.86
Total1400.23 60.81
xuy = 309 mm               Puc = C1.fck.B.D
ku = 1.031
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.027
C1 = 0.446 x (1 - C3/6) = 0.370
Puc = 0.370 x 20.00 x 450 x 300
= 998.00 kN
Puy1 = Puc + Pus(Total)
= 998.00 + (1400.23)
= 2398.23 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.426
Muc = 998.00 x (0.5 x 300 - 0.426 x 300) = 22.09 kN-m
Muy1 = Muc + Mus(Total)
= 22.09 + (60.81)
= 82.90 kN-m
Pu/Puz = 0.798
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.798, an = 1.997
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((49.48/110.75)1.997) + ((47.80/82.90)1.997)
= 0.533 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 25/4 = 6 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (6, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 25 = 400 mm
               iii) 300 mm
               Required spacing = minimum of (300, 400, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG26 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #25 - 12 nos. (5890 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 3256.55 kN
MomentX,(Mx) = 10.46 kN-m
MomentY,(My) = 10.34 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C340.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 67.411 kN.m
My_MinEccen = 65.131 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 67.411 kN.m
My = max(My,My_MinEccen) + MuaddY = 65.131 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 10.46/3256.55 = 3 mm
Actual eccenY = My / P = 10.34/3256.55 = 3 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(3,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(3,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #32 - 12 nos. (9651 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1608 0.00304 354.24 8.92 345.32 555.45 169 93.87
2 1608 0.00248 344.68 8.92 335.76 540.07 101 54.76
3 1608 0.00193 325.05 8.91 316.14 508.52 34 17.19
4 1608 0.00137 274.13 8.04 266.10 428.01 -34 -14.47
5 1608 0.00081 162.89 5.79 157.11 252.71 -101 -25.62
6 1608 0.00026 51.65 2.16 49.50 79.62 -169 -13.46
Total2364.37 112.27
xux = 425 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 425/450 = 0.340
Puc = 0.340 x 20.00 x 300 x 450
= 918.84 kN
Pux1 = Puc + Pus(Total)
= 918.84 + (2364.37)
= 3283.22 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 425/450
= 0.393
Muc = 918.84 x (0.5 x 450 - 0.393 x 450) = 44.14 kN-m
Mux1 = Muc + Mus(Total)
= 44.14 + (112.27)
= 156.41 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 4825 0.00282 352.36 8.92 343.44 1657.27 94 155.78
2 4825 0.00069 137.84 5.09 132.75 640.57 -94 -60.21
Total2297.84 95.57
xuy = 305 mm               Puc = C1.fck.B.D
ku = 1.016
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.083
C1 = 0.446 x (1 - C3/6) = 0.366
Puc = 0.366 x 20.00 x 450 x 300
= 986.88 kN
Puy1 = Puc + Pus(Total)
= 986.88 + (2297.84)
= 3284.72 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.421
Muc = 986.88 x (0.5 x 300 - 0.421 x 300) = 23.28 kN-m
Muy1 = Muc + Mus(Total)
= 23.28 + (95.57)
= 118.85 kN-m
Pu/Puz = 0.788
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.788, an = 1.980
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((67.41/156.41)1.980) + ((65.13/118.85)1.980)
= 0.493 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 32/4 = 8 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (8, 6) = 8 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 32 = 512 mm
               iii) 300 mm
               Required spacing = minimum of (300, 512, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG26 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #32 - 12 nos. (9651 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 600 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 4203.35 kN
MomentX,(Mx) = 3.79 kN-m
MomentY,(My) = 6.84 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C340.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 600 = 5.63 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 112.440 kN.m
My_MinEccen = 84.067 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 112.440 kN.m
My = max(My,My_MinEccen) + MuaddY = 84.067 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 3.79/4203.35 = 1 mm
Actual eccenY = My / P = 6.84/4203.35 = 2 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (600/30) = 27 mm
eccenXMin = max(27,20)= 27 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 27 mm
eccenX = max(Actual eccenX,eccenXMin) = max(1,27) = 27 mm
eccenY = max(Actual eccenY,eccenYMin) = max(2,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(27 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 600(30 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #32 - 14 nos. (11259 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1608 0.00314 355.14 8.92 346.22 556.90 244 135.88
2 1608 0.00268 349.75 8.92 340.83 548.22 163 89.18
3 1608 0.00222 335.76 8.92 326.84 525.73 81 42.76
4 1608 0.00176 314.60 8.79 305.82 491.91 0 0.00
5 1608 0.00129 258.97 7.81 251.16 403.98 -81 -32.86
6 1608 0.00083 166.60 5.88 160.72 258.52 -163 -42.05
7 1608 0.00037 74.24 3.00 71.24 114.58 -244 -27.96
Total2899.84 164.95
xux = 609 mm               Puc = C1.fck.B.D
ku = 1.016
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.083
C1 = 0.446 x (1 - C3/6) = 0.366
Puc = 0.366 x 20.00 x 300 x 600
= 1315.83 kN
Pux1 = Puc + Pus(Total)
= 1315.83 + (2899.84)
= 4215.68 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.421
Muc = 1315.83 x (0.5 x 600 - 0.421 x 600) = 62.09 kN-m
Mux1 = Muc + Mus(Total)
= 62.09 + (164.95)
= 227.04 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 5630 0.00273 350.96 8.92 342.04 1925.59 94 181.01
2 5630 0.00084 168.63 5.94 162.69 915.91 -94 -86.10
Total2841.50 94.91
xuy = 328 mm               Puc = C1.fck.B.D
ku = 1.094
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.843
C1 = 0.446 x (1 - C3/6) = 0.383
Puc = 0.383 x 20.00 x 600 x 300
= 1379.90 kN
Puy1 = Puc + Pus(Total)
= 1379.90 + (2841.50)
= 4221.40 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.442
Muc = 1379.90 x (0.5 x 300 - 0.442 x 300) = 24.18 kN-m
Muy1 = Muc + Mus(Total)
= 24.18 + (94.91)
= 119.09 kN-m
Pu/Puz = 0.837
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.837, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((112.44/227.04)2.000) + ((84.07/119.09)2.000)
= 0.744 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 32/4 = 8 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (8, 6) = 8 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 32 = 512 mm
               iii) 300 mm
               Required spacing = minimum of (300, 512, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG26 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 600 mm
Provide #32 - 14 nos. (11259 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG27 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 628.41 kN
MomentX,(Mx) = 46.45 kN-m
MomentY,(My) = 42.74 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C350.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 13.008 kN.m
My_MinEccen = 12.568 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 46.454 kN.m
My = max(My,My_MinEccen) + MuaddY = 42.742 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 46.45/628.41 = 74 mm
Actual eccenY = My / P = 42.74/628.41 = 68 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(74,21) = 74 mm
eccenY = max(Actual eccenY,eccenYMin) = max(68,20) = 68 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(74 mm) > 0.05 x 300(15 mm)
and eccenY(68 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00289 352.91 8.92 343.99 77.81 179 13.93
2 226 0.00170 310.78 8.71 302.07 68.33 90 6.12
3 226 0.00050 100.67 3.92 96.74 21.88 0 0.00
4 226 -0.00069 -137.73 0.00 -137.73 -31.15 -90 2.79
5 226 -0.00188 -322.34 0.00 -322.34 -72.91 -179 13.05
Total63.95 35.88
xux = 263 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 263/450 = 0.210
Puc = 0.210 x 20.00 x 300 x 450
= 567.63 kN
Pux1 = Puc + Pus(Total)
= 567.63 + (63.95)
= 631.58 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 263/450
= 0.243
Muc = 567.63 x (0.5 x 450 - 0.243 x 450) = 65.66 kN-m
Mux1 = Muc + Mus(Total)
= 65.66 + (35.88)
= 101.55 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00262 348.18 8.92 339.26 191.85 104 19.95
2 565 -0.00136 -272.58 0.00 -272.58 -154.14 -104 16.03
Total37.71 35.98
xuy = 183 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 183/300 = 0.219
Puc = 0.219 x 20.00 x 450 x 300
= 592.31 kN
Puy1 = Puc + Pus(Total)
= 592.31 + (37.71)
= 630.02 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 183/300
= 0.254
Muc = 592.31 x (0.5 x 300 - 0.254 x 300) = 43.80 kN-m
Muy1 = Muc + Mus(Total)
= 43.80 + (35.98)
= 79.78 kN-m
Pu/Puz = 0.404
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.404, an = 1.339
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((46.45/101.55)1.339) + ((42.74/79.78)1.339)
= 0.784 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG27 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1259.29 kN
MomentX,(Mx) = 39.57 kN-m
MomentY,(My) = 41.12 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C350.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 26.067 kN.m
My_MinEccen = 25.186 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 39.574 kN.m
My = max(My,My_MinEccen) + MuaddY = 41.116 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 39.57/1259.29 = 31 mm
Actual eccenY = My / P = 41.12/1259.29 = 33 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(31,21) = 31 mm
eccenY = max(Actual eccenY,eccenYMin) = max(33,20) = 33 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(31 mm) > 0.05 x 300(15 mm)
and eccenY(33 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 4 nos. + #16 - 6 nos. (1659 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00309 354.70 8.92 345.78 139.05 177 24.61
2 226 0.00234 340.17 8.92 331.25 74.93 89 6.63
3 402 0.00159 302.48 8.54 293.94 118.20 0 0.00
4 226 0.00083 166.48 5.88 160.60 36.33 -89 -3.22
5 402 0.00008 15.87 0.69 15.18 6.10 -177 -1.08
Total374.61 26.95
xux = 411 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 411/450 = 0.329
Puc = 0.329 x 20.00 x 300 x 450
= 888.47 kN
Pux1 = Puc + Pus(Total)
= 888.47 + (374.61)
= 1263.08 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 411/450
= 0.380
Muc = 888.47 x (0.5 x 450 - 0.380 x 450) = 47.88 kN-m
Mux1 = Muc + Mus(Total)
= 47.88 + (26.95)
= 74.82 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 829 0.00291 353.09 8.92 344.17 285.45 102 29.12
2 829 0.00039 77.98 3.14 74.84 62.07 -102 -6.33
Total347.52 22.78
xuy = 284 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 284/300 = 0.340
Puc = 0.340 x 20.00 x 450 x 300
= 918.84 kN
Puy1 = Puc + Pus(Total)
= 918.84 + (347.52)
= 1266.37 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 284/300
= 0.393
Muc = 918.84 x (0.5 x 300 - 0.393 x 300) = 29.43 kN-m
Muy1 = Muc + Mus(Total)
= 29.43 + (22.78)
= 52.21 kN-m
Pu/Puz = 0.734
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.734, an = 1.889
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((39.57/74.82)1.889) + ((41.12/52.21)1.889)
= 0.937 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG27 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 4 nos. + #16 - 6 nos. (1659 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1899.61 kN
MomentX,(Mx) = 35.04 kN-m
MomentY,(My) = 39.76 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C350.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 39.322 kN.m
My_MinEccen = 37.992 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 39.322 kN.m
My = max(My,My_MinEccen) + MuaddY = 39.758 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 35.04/1899.61 = 18 mm
Actual eccenY = My / P = 39.76/1899.61 = 21 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(18,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(21,20) = 21 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(21 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 18 nos. (3619 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00313 355.01 8.92 346.09 139.17 177 24.63
2 402 0.00278 352.00 8.92 343.08 137.96 133 18.31
3 402 0.00244 343.53 8.92 334.61 134.55 89 11.91
4 402 0.00209 331.20 8.92 322.28 129.60 44 5.73
5 402 0.00175 314.19 8.78 305.41 122.81 0 0.00
6 402 0.00141 281.17 8.13 273.03 109.79 -44 -4.86
7 402 0.00106 212.33 6.96 205.38 82.59 -89 -7.31
8 402 0.00072 143.50 5.25 138.25 55.59 -133 -7.38
9 402 0.00037 74.67 3.02 71.65 28.81 -177 -5.10
Total940.88 35.94
xux = 450 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 450/450 = 0.360
Puc = 0.360 x 20.00 x 300 x 450
= 972.00 kN
Pux1 = Puc + Pus(Total)
= 972.00 + (940.88)
= 1912.88 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 450/450
= 0.416
Muc = 972.00 x (0.5 x 450 - 0.416 x 450) = 36.74 kN-m
Mux1 = Muc + Mus(Total)
= 36.74 + (35.94)
= 72.68 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1810 0.00283 352.39 8.92 343.47 621.52 102 63.40
2 1810 0.00073 146.64 5.34 141.30 255.69 -102 -26.08
Total877.21 37.32
xuy = 323 mm               Puc = C1.fck.B.D
ku = 1.078
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.884
C1 = 0.446 x (1 - C3/6) = 0.380
Puc = 0.380 x 20.00 x 450 x 300
= 1026.69 kN
Puy1 = Puc + Pus(Total)
= 1026.69 + (877.21)
= 1903.89 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.438
Muc = 1026.69 x (0.5 x 300 - 0.438 x 300) = 19.02 kN-m
Muy1 = Muc + Mus(Total)
= 19.02 + (37.32)
= 56.33 kN-m
Pu/Puz = 0.823
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.823, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((39.32/72.68)2.000) + ((39.76/56.33)2.000)
= 0.791 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG27 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 18 nos. (3619 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 2563.04 kN
MomentX,(Mx) = 31.53 kN-m
MomentY,(My) = 37.58 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C350.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 53.055 kN.m
My_MinEccen = 51.261 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 53.055 kN.m
My = max(My,My_MinEccen) + MuaddY = 51.261 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 31.53/2563.04 = 12 mm
Actual eccenY = My / P = 37.58/2563.04 = 15 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(12,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(15,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #25 - 14 nos. (6872 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 982 0.00306 354.46 8.92 345.54 339.24 173 58.52
2 982 0.00259 347.36 8.92 338.44 332.26 115 38.21
3 982 0.00211 331.79 8.92 322.87 316.98 58 18.23
4 982 0.00163 306.91 8.62 298.29 292.84 0 0.00
5 982 0.00116 231.26 7.33 223.93 219.84 -58 -12.64
6 982 0.00068 135.85 5.03 130.82 128.43 -115 -14.77
7 982 0.00020 40.44 1.71 38.73 38.02 -173 -6.56
Total1667.62 80.99
xux = 422 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 422/450 = 0.338
Puc = 0.338 x 20.00 x 300 x 450
= 911.25 kN
Pux1 = Puc + Pus(Total)
= 911.25 + (1667.62)
= 2578.87 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 422/450
= 0.390
Muc = 911.25 x (0.5 x 450 - 0.390 x 450) = 45.11 kN-m
Mux1 = Muc + Mus(Total)
= 45.11 + (80.99)
= 126.09 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 3436 0.00288 352.81 8.92 343.89 1181.65 98 115.21
2 3436 0.00063 126.24 4.74 121.50 417.49 -98 -40.71
Total1599.14 74.51
xuy = 302 mm               Puc = C1.fck.B.D
ku = 1.008
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.112
C1 = 0.446 x (1 - C3/6) = 0.363
Puc = 0.363 x 20.00 x 450 x 300
= 980.97 kN
Puy1 = Puc + Pus(Total)
= 980.97 + (1599.14)
= 2580.11 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.419
Muc = 980.97 x (0.5 x 300 - 0.419 x 300) = 23.92 kN-m
Muy1 = Muc + Mus(Total)
= 23.92 + (74.51)
= 98.42 kN-m
Pu/Puz = 0.779
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.779, an = 1.964
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((53.05/126.09)1.964) + ((51.26/98.42)1.964)
= 0.460 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 25/4 = 6 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (6, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 25 = 400 mm
               iii) 300 mm
               Required spacing = minimum of (300, 400, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG27 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #25 - 14 nos. (6872 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 3260.92 kN
MomentX,(Mx) = 18.71 kN-m
MomentY,(My) = 24.33 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C350.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 70.925 kN.m
My_MinEccen = 65.218 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 70.925 kN.m
My = max(My,My_MinEccen) + MuaddY = 65.218 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 18.71/3260.92 = 6 mm
Actual eccenY = My / P = 24.33/3260.92 = 7 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(6,22) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(7,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #32 - 12 nos. (9651 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1608 0.00304 354.24 8.92 345.32 555.45 169 93.87
2 1608 0.00248 344.68 8.92 335.76 540.07 101 54.76
3 1608 0.00193 325.05 8.91 316.14 508.52 34 17.19
4 1608 0.00137 274.13 8.04 266.10 428.01 -34 -14.47
5 1608 0.00081 162.89 5.79 157.11 252.71 -101 -25.62
6 1608 0.00026 51.65 2.16 49.50 79.62 -169 -13.46
Total2364.37 112.27
xux = 425 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 425/450 = 0.340
Puc = 0.340 x 20.00 x 300 x 450
= 918.84 kN
Pux1 = Puc + Pus(Total)
= 918.84 + (2364.37)
= 3283.22 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 425/450
= 0.393
Muc = 918.84 x (0.5 x 450 - 0.393 x 450) = 44.14 kN-m
Mux1 = Muc + Mus(Total)
= 44.14 + (112.27)
= 156.41 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 4825 0.00282 352.36 8.92 343.44 1657.27 94 155.78
2 4825 0.00069 137.84 5.09 132.75 640.57 -94 -60.21
Total2297.84 95.57
xuy = 305 mm               Puc = C1.fck.B.D
ku = 1.016
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.083
C1 = 0.446 x (1 - C3/6) = 0.366
Puc = 0.366 x 20.00 x 450 x 300
= 986.88 kN
Puy1 = Puc + Pus(Total)
= 986.88 + (2297.84)
= 3284.72 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.421
Muc = 986.88 x (0.5 x 300 - 0.421 x 300) = 23.28 kN-m
Muy1 = Muc + Mus(Total)
= 23.28 + (95.57)
= 118.85 kN-m
Pu/Puz = 0.789
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.789, an = 1.982
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((70.92/156.41)1.982) + ((65.22/118.85)1.982)
= 0.513 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 32/4 = 8 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (8, 6) = 8 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 32 = 512 mm
               iii) 300 mm
               Required spacing = minimum of (300, 512, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG27 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #32 - 12 nos. (9651 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG28 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 598.62 kN
MomentX,(Mx) = 11.73 kN-m
MomentY,(My) = 17.27 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C360.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 12.391 kN.m
My_MinEccen = 11.972 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 12.391 kN.m
My = max(My,My_MinEccen) + MuaddY = 17.271 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 11.73/598.62 = 20 mm
Actual eccenY = My / P = 17.27/598.62 = 29 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(20,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(29,20) = 29 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(29 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00287 352.75 8.92 343.83 77.77 179 13.92
2 226 0.00164 307.28 8.63 298.65 67.55 90 6.05
3 226 0.00041 82.07 3.28 78.78 17.82 0 0.00
4 226 -0.00082 -163.73 0.00 -163.73 -37.03 -90 3.31
5 226 -0.00205 -329.49 0.00 -329.49 -74.53 -179 13.34
Total51.58 36.62
xux = 255 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 255/450 = 0.204
Puc = 0.204 x 20.00 x 300 x 450
= 550.55 kN
Pux1 = Puc + Pus(Total)
= 550.55 + (51.58)
= 602.13 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 255/450
= 0.236
Muc = 550.55 x (0.5 x 450 - 0.236 x 450) = 65.50 kN-m
Mux1 = Muc + Mus(Total)
= 65.50 + (36.62)
= 102.12 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00260 347.59 8.92 338.67 191.51 104 19.92
2 565 -0.00149 -293.52 0.00 -293.52 -165.98 -104 17.26
Total25.53 37.18
xuy = 178 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 178/300 = 0.214
Puc = 0.214 x 20.00 x 450 x 300
= 577.13 kN
Puy1 = Puc + Pus(Total)
= 577.13 + (25.53)
= 602.65 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 178/300
= 0.247
Muc = 577.13 x (0.5 x 300 - 0.247 x 300) = 43.80 kN-m
Muy1 = Muc + Mus(Total)
= 43.80 + (37.18)
= 80.98 kN-m
Pu/Puz = 0.385
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.385, an = 1.308
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((12.39/102.12)1.308) + ((17.27/80.98)1.308)
= 0.196 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG28 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1195.36 kN
MomentX,(Mx) = 7.17 kN-m
MomentY,(My) = 16.88 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C360.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 24.744 kN.m
My_MinEccen = 23.907 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 24.744 kN.m
My = max(My,My_MinEccen) + MuaddY = 23.907 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 7.17/1195.36 = 6 mm
Actual eccenY = My / P = 16.88/1195.36 = 14 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(6,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(14,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00312 354.99 8.92 346.07 78.28 179 14.01
2 226 0.00239 342.22 8.92 333.30 75.39 90 6.75
3 226 0.00166 308.82 8.67 300.15 67.89 0 0.00
4 226 0.00093 186.72 6.38 180.33 40.79 -90 -3.65
5 226 0.00020 40.65 1.72 38.93 8.81 -179 -1.58
Total271.16 15.53
xux = 429 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 429/450 = 0.343
Puc = 0.343 x 20.00 x 300 x 450
= 926.44 kN
Pux1 = Puc + Pus(Total)
= 926.44 + (271.16)
= 1197.60 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 429/450
= 0.397
Muc = 926.44 x (0.5 x 450 - 0.397 x 450) = 43.15 kN-m
Mux1 = Muc + Mus(Total)
= 43.15 + (15.53)
= 58.68 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00295 353.50 8.92 344.58 194.86 104 20.27
2 565 0.00049 97.93 3.83 94.09 53.21 -104 -5.53
Total248.07 14.73
xuy = 295 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 295/300 = 0.354
Puc = 0.354 x 20.00 x 450 x 300
= 956.81 kN
Puy1 = Puc + Pus(Total)
= 956.81 + (248.07)
= 1204.88 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 295/300
= 0.410
Muc = 956.81 x (0.5 x 300 - 0.410 x 300) = 25.98 kN-m
Muy1 = Muc + Mus(Total)
= 25.98 + (14.73)
= 40.71 kN-m
Pu/Puz = 0.768
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.768, an = 1.946
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((24.74/58.68)1.946) + ((23.91/40.71)1.946)
= 0.541 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG28 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1793.71 kN
MomentX,(Mx) = 4.19 kN-m
MomentY,(My) = 18.99 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C360.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 37.130 kN.m
My_MinEccen = 35.874 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 37.130 kN.m
My = max(My,My_MinEccen) + MuaddY = 35.874 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 4.19/1793.71 = 2 mm
Actual eccenY = My / P = 18.99/1793.71 = 11 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(2,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(11,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 16 nos. (3217 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00313 355.01 8.92 346.09 139.17 177 24.63
2 402 0.00273 351.11 8.92 342.19 137.60 126 17.40
3 402 0.00234 340.23 8.92 331.31 133.23 76 10.11
4 402 0.00195 325.78 8.91 316.87 127.42 25 3.22
5 402 0.00155 299.44 8.48 290.96 117.00 -25 -2.96
6 402 0.00116 232.00 7.35 224.65 90.34 -76 -6.85
7 402 0.00077 153.33 5.53 147.81 59.44 -126 -7.51
8 402 0.00037 74.67 3.02 71.65 28.81 -177 -5.10
Total833.01 32.93
xux = 450 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 450/450 = 0.360
Puc = 0.360 x 20.00 x 300 x 450
= 972.00 kN
Pux1 = Puc + Pus(Total)
= 972.00 + (833.01)
= 1805.01 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 450/450
= 0.416
Muc = 972.00 x (0.5 x 450 - 0.416 x 450) = 36.74 kN-m
Mux1 = Muc + Mus(Total)
= 36.74 + (32.93)
= 69.67 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1608 0.00283 352.39 8.92 343.47 552.46 102 56.35
2 1608 0.00073 146.64 5.34 141.30 227.28 -102 -23.18
Total779.74 33.17
xuy = 323 mm               Puc = C1.fck.B.D
ku = 1.078
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.884
C1 = 0.446 x (1 - C3/6) = 0.380
Puc = 0.380 x 20.00 x 450 x 300
= 1026.69 kN
Puy1 = Puc + Pus(Total)
= 1026.69 + (779.74)
= 1806.43 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.438
Muc = 1026.69 x (0.5 x 300 - 0.438 x 300) = 19.02 kN-m
Muy1 = Muc + Mus(Total)
= 19.02 + (33.17)
= 52.19 kN-m
Pu/Puz = 0.820
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.820, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((37.13/69.67)2.000) + ((35.87/52.19)2.000)
= 0.757 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG28 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 16 nos. (3217 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 2391.06 kN
MomentX,(Mx) = 2.92 kN-m
MomentY,(My) = 22.68 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C360.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 49.495 kN.m
My_MinEccen = 47.821 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 49.495 kN.m
My = max(My,My_MinEccen) + MuaddY = 47.821 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 2.92/2391.06 = 1 mm
Actual eccenY = My / P = 22.68/2391.06 = 9 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(1,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(9,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #25 - 12 nos. (5890 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 982 0.00308 354.56 8.92 345.64 339.33 173 58.53
2 982 0.00252 345.54 8.92 336.62 330.48 104 34.20
3 982 0.00196 326.20 8.92 317.28 311.49 35 10.75
4 982 0.00140 279.92 8.12 271.81 266.85 -35 -9.21
5 982 0.00084 168.23 5.93 162.30 159.34 -104 -16.49
6 982 0.00028 56.53 2.34 54.19 53.20 -173 -9.18
Total1460.68 68.61
xux = 432 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 432/450 = 0.346
Puc = 0.346 x 20.00 x 300 x 450
= 934.03 kN
Pux1 = Puc + Pus(Total)
= 934.03 + (1460.68)
= 2394.71 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 432/450
= 0.400
Muc = 934.03 x (0.5 x 450 - 0.400 x 450) = 42.14 kN-m
Mux1 = Muc + Mus(Total)
= 42.14 + (68.61)
= 110.75 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 2945 0.00284 352.51 8.92 343.59 1011.96 98 98.67
2 2945 0.00068 136.89 5.06 131.83 388.27 -98 -37.86
Total1400.23 60.81
xuy = 309 mm               Puc = C1.fck.B.D
ku = 1.031
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.027
C1 = 0.446 x (1 - C3/6) = 0.370
Puc = 0.370 x 20.00 x 450 x 300
= 998.00 kN
Puy1 = Puc + Pus(Total)
= 998.00 + (1400.23)
= 2398.23 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.426
Muc = 998.00 x (0.5 x 300 - 0.426 x 300) = 22.09 kN-m
Muy1 = Muc + Mus(Total)
= 22.09 + (60.81)
= 82.90 kN-m
Pu/Puz = 0.798
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.798, an = 1.997
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((49.49/110.75)1.997) + ((47.82/82.90)1.997)
= 0.533 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 25/4 = 6 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (6, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 25 = 400 mm
               iii) 300 mm
               Required spacing = minimum of (300, 400, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG28 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #25 - 12 nos. (5890 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 2980.01 kN
MomentX,(Mx) = 0.77 kN-m
MomentY,(My) = 15.81 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C360.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 64.815 kN.m
My_MinEccen = 59.600 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 64.815 kN.m
My = max(My,My_MinEccen) + MuaddY = 59.600 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 0.77/2980.01 = 0 mm
Actual eccenY = My / P = 15.81/2980.01 = 5 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(0,22) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(5,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #32 - 10 nos. (8042 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1608 0.00306 354.40 8.92 345.48 555.71 169 93.91
2 1608 0.00239 342.06 8.92 333.14 535.86 85 45.28
3 1608 0.00172 312.46 8.75 303.71 488.51 0 0.00
4 1608 0.00105 210.91 6.93 203.99 328.11 -85 -27.73
5 1608 0.00039 77.38 3.12 74.27 119.46 -169 -20.19
Total2027.64 91.28
xux = 443 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 443/450 = 0.354
Puc = 0.354 x 20.00 x 300 x 450
= 956.81 kN
Pux1 = Puc + Pus(Total)
= 956.81 + (2027.64)
= 2984.46 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 443/450
= 0.410
Muc = 956.81 x (0.5 x 450 - 0.410 x 450) = 38.97 kN-m
Mux1 = Muc + Mus(Total)
= 38.97 + (91.28)
= 130.25 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 4021 0.00276 351.83 8.92 342.91 1378.91 94 129.62
2 4021 0.00079 157.22 5.63 151.59 609.57 -94 -57.30
Total1988.48 72.32
xuy = 319 mm               Puc = C1.fck.B.D
ku = 1.063
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.929
C1 = 0.446 x (1 - C3/6) = 0.377
Puc = 0.377 x 20.00 x 450 x 300
= 1017.83 kN
Puy1 = Puc + Pus(Total)
= 1017.83 + (1988.48)
= 3006.31 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.435
Muc = 1017.83 x (0.5 x 300 - 0.435 x 300) = 19.97 kN-m
Muy1 = Muc + Mus(Total)
= 19.97 + (72.32)
= 92.29 kN-m
Pu/Puz = 0.817
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.817, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((64.82/130.25)2.000) + ((59.60/92.29)2.000)
= 0.665 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 32/4 = 8 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (8, 6) = 8 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 32 = 512 mm
               iii) 300 mm
               Required spacing = minimum of (300, 512, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG28 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #32 - 10 nos. (8042 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG29 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 387.35 kN
MomentX,(Mx) = 10.84 kN-m
MomentY,(My) = 79.05 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C370.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 8.018 kN.m
My_MinEccen = 7.747 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 10.838 kN.m
My = max(My,My_MinEccen) + MuaddY = 79.046 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 10.84/387.35 = 28 mm
Actual eccenY = My / P = 79.05/387.35 = 204 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(28,21) = 28 mm
eccenY = max(Actual eccenY,eccenYMin) = max(204,20) = 204 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(28 mm) > 0.05 x 300(15 mm)
and eccenY(204 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 12 nos. (1357 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00271 350.44 8.92 341.52 77.25 179 13.83
2 226 0.00147 291.80 8.30 283.50 64.13 107 6.89
3 226 0.00024 47.67 2.00 45.67 10.33 36 0.37
4 226 -0.00100 -199.19 0.00 -199.19 -45.06 -36 1.61
5 226 -0.00223 -336.20 0.00 -336.20 -76.05 -107 8.17
6 226 -0.00346 -357.97 0.00 -357.97 -80.97 -179 14.49
Total-50.37 45.36
xux = 203 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 203/450 = 0.162
Puc = 0.162 x 20.00 x 300 x 450
= 438.54 kN
Pux1 = Puc + Pus(Total)
= 438.54 + (-50.37)
= 388.17 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 203/450
= 0.188
Muc = 438.54 x (0.5 x 450 - 0.188 x 450) = 61.63 kN-m
Mux1 = Muc + Mus(Total)
= 61.63 + (45.36)
= 106.99 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 679 0.00223 336.11 8.92 327.19 222.03 104 23.09
2 679 -0.00352 -358.49 0.00 -358.49 -243.26 -104 25.30
Total-21.24 48.39
xuy = 127 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 127/300 = 0.152
Puc = 0.152 x 20.00 x 450 x 300
= 410.06 kN
Puy1 = Puc + Pus(Total)
= 410.06 + (-21.24)
= 388.83 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 127/300
= 0.176
Muc = 410.06 x (0.5 x 300 - 0.176 x 300) = 39.92 kN-m
Muy1 = Muc + Mus(Total)
= 39.92 + (48.39)
= 88.31 kN-m
Pu/Puz = 0.238
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.238, an = 1.064
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((10.84/106.99)1.064) + ((79.05/88.31)1.064)
= 0.976 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG29 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 12 nos. (1357 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 795.00 kN
MomentX,(Mx) = 9.00 kN-m
MomentY,(My) = 72.81 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C370.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 16.456 kN.m
My_MinEccen = 15.900 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 16.456 kN.m
My = max(My,My_MinEccen) + MuaddY = 72.809 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 9.00/795.00 = 11 mm
Actual eccenY = My / P = 72.81/795.00 = 92 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(11,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(92,20) = 92 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(92 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 12 nos. (1357 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00296 353.56 8.92 344.64 77.96 179 13.95
2 226 0.00212 332.24 8.92 323.32 73.13 107 7.85
3 226 0.00128 256.80 7.78 249.03 56.33 36 2.02
4 226 0.00045 89.08 3.53 85.55 19.35 -36 -0.69
5 226 -0.00039 -78.64 0.00 -78.64 -17.79 -107 1.91
6 226 -0.00123 -246.36 0.00 -246.36 -55.73 -179 9.97
Total153.25 35.02
xux = 299 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 299/450 = 0.239
Puc = 0.239 x 20.00 x 300 x 450
= 645.47 kN
Pux1 = Puc + Pus(Total)
= 645.47 + (153.25)
= 798.72 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 299/450
= 0.276
Muc = 645.47 x (0.5 x 450 - 0.276 x 450) = 64.99 kN-m
Mux1 = Muc + Mus(Total)
= 64.99 + (35.02)
= 100.01 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 679 0.00272 350.87 8.92 341.95 232.04 104 24.13
2 679 -0.00079 -157.19 0.00 -157.19 -106.67 -104 11.09
Total125.37 35.23
xuy = 207 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 207/300 = 0.249
Puc = 0.249 x 20.00 x 450 x 300
= 672.05 kN
Puy1 = Puc + Pus(Total)
= 672.05 + (125.37)
= 797.42 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 207/300
= 0.288
Muc = 672.05 x (0.5 x 300 - 0.288 x 300) = 42.82 kN-m
Muy1 = Muc + Mus(Total)
= 42.82 + (35.23)
= 78.04 kN-m
Pu/Puz = 0.489
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.489, an = 1.482
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((16.46/100.01)1.482) + ((72.81/78.04)1.482)
= 0.971 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG29 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 12 nos. (1357 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1202.38 kN
MomentX,(Mx) = 7.54 kN-m
MomentY,(My) = 70.52 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C370.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 24.889 kN.m
My_MinEccen = 24.048 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 24.889 kN.m
My = max(My,My_MinEccen) + MuaddY = 70.519 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 7.54/1202.38 = 6 mm
Actual eccenY = My / P = 70.52/1202.38 = 59 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(6,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(59,20) = 59 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(59 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 12 nos. (2413 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00303 354.14 8.92 345.22 138.82 177 24.57
2 402 0.00233 339.82 8.92 330.90 133.06 106 14.13
3 402 0.00163 306.77 8.62 298.15 119.89 35 4.24
4 402 0.00093 186.65 6.38 180.27 72.49 -35 -2.57
5 402 0.00024 47.07 1.98 45.10 18.13 -106 -1.93
6 402 -0.00046 -92.50 0.00 -92.50 -37.20 -177 6.58
Total445.20 45.04
xux = 355 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 355/450 = 0.284
Puc = 0.284 x 20.00 x 300 x 450
= 766.97 kN
Pux1 = Puc + Pus(Total)
= 766.97 + (445.20)
= 1212.17 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 355/450
= 0.328
Muc = 766.97 x (0.5 x 450 - 0.328 x 450) = 59.28 kN-m
Mux1 = Muc + Mus(Total)
= 59.28 + (45.04)
= 104.32 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1206 0.00282 352.36 8.92 343.44 414.31 102 42.26
2 1206 -0.00005 -10.04 0.00 -10.04 -12.11 -102 1.24
Total402.20 43.50
xuy = 248 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 248/300 = 0.298
Puc = 0.298 x 20.00 x 450 x 300
= 804.94 kN
Puy1 = Puc + Pus(Total)
= 804.94 + (402.20)
= 1207.14 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 248/300
= 0.345
Muc = 804.94 x (0.5 x 300 - 0.345 x 300) = 37.55 kN-m
Muy1 = Muc + Mus(Total)
= 37.55 + (43.50)
= 81.05 kN-m
Pu/Puz = 0.618
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.618, an = 1.697
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((24.89/104.32)1.697) + ((70.52/81.05)1.697)
= 0.877 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG29 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 12 nos. (2413 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1610.17 kN
MomentX,(Mx) = 5.58 kN-m
MomentY,(My) = 67.40 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C370.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 33.330 kN.m
My_MinEccen = 32.203 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 33.330 kN.m
My = max(My,My_MinEccen) + MuaddY = 67.395 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 5.58/1610.17 = 3 mm
Actual eccenY = My / P = 67.40/1610.17 = 42 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(3,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(42,20) = 42 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(42 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 18 nos. (3619 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00307 354.47 8.92 345.55 138.96 177 24.60
2 402 0.00267 349.36 8.92 340.44 136.90 133 18.17
3 402 0.00226 337.46 8.92 328.54 132.11 89 11.69
4 402 0.00186 321.31 8.88 312.43 125.64 44 5.56
5 402 0.00146 290.94 8.28 282.66 113.66 0 0.00
6 402 0.00106 212.63 6.96 205.67 82.70 -44 -3.66
7 402 0.00066 132.53 4.93 127.60 51.31 -89 -4.54
8 402 0.00026 52.44 2.19 50.25 20.21 -133 -2.68
9 402 -0.00014 -27.66 0.00 -27.66 -11.12 -177 1.97
Total790.37 51.11
xux = 387 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 387/450 = 0.309
Puc = 0.309 x 20.00 x 300 x 450
= 835.31 kN
Pux1 = Puc + Pus(Total)
= 835.31 + (790.37)
= 1625.68 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 387/450
= 0.358
Muc = 835.31 x (0.5 x 450 - 0.358 x 450) = 53.56 kN-m
Mux1 = Muc + Mus(Total)
= 53.56 + (51.11)
= 104.67 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1810 0.00289 352.94 8.92 344.02 622.52 102 63.50
2 1810 0.00030 59.46 2.45 57.00 103.15 -102 -10.52
Total725.66 52.98
xuy = 275 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 275/300 = 0.330
Puc = 0.330 x 20.00 x 450 x 300
= 892.27 kN
Puy1 = Puc + Pus(Total)
= 892.27 + (725.66)
= 1617.93 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 275/300
= 0.382
Muc = 892.27 x (0.5 x 300 - 0.382 x 300) = 31.62 kN-m
Muy1 = Muc + Mus(Total)
= 31.62 + (52.98)
= 84.60 kN-m
Pu/Puz = 0.697
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.697, an = 1.829
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((33.33/104.67)1.829) + ((67.40/84.60)1.829)
= 0.783 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG29 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 18 nos. (3619 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 2016.95 kN
MomentX,(Mx) = 1.83 kN-m
MomentY,(My) = 44.81 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C370.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 43.869 kN.m
My_MinEccen = 40.339 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 43.869 kN.m
My = max(My,My_MinEccen) + MuaddY = 44.810 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 1.83/2016.95 = 1 mm
Actual eccenY = My / P = 44.81/2016.95 = 22 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(1,22) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(22,20) = 22 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(22 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 14 nos. (4398 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00310 354.73 8.92 345.81 217.28 175 38.02
2 628 0.00262 348.28 8.92 339.36 213.23 117 24.88
3 628 0.00215 333.29 8.92 324.37 203.81 58 11.89
4 628 0.00168 309.75 8.69 301.06 189.16 0 0.00
5 628 0.00121 241.34 7.52 233.83 146.92 -58 -8.57
6 628 0.00073 146.91 5.35 141.56 88.95 -117 -10.38
7 628 0.00026 52.48 2.19 50.30 31.60 -175 -5.53
Total1090.94 50.31
xux = 432 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 432/450 = 0.346
Puc = 0.346 x 20.00 x 300 x 450
= 934.03 kN
Pux1 = Puc + Pus(Total)
= 934.03 + (1090.94)
= 2024.97 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 432/450
= 0.400
Muc = 934.03 x (0.5 x 450 - 0.400 x 450) = 42.14 kN-m
Mux1 = Muc + Mus(Total)
= 42.14 + (50.31)
= 92.45 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 2199 0.00287 352.75 8.92 343.83 756.13 100 75.61
2 2199 0.00066 131.36 4.90 126.46 278.10 -100 -27.81
Total1034.24 47.80
xuy = 309 mm               Puc = C1.fck.B.D
ku = 1.031
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.027
C1 = 0.446 x (1 - C3/6) = 0.370
Puc = 0.370 x 20.00 x 450 x 300
= 998.00 kN
Puy1 = Puc + Pus(Total)
= 998.00 + (1034.24)
= 2032.23 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.426
Muc = 998.00 x (0.5 x 300 - 0.426 x 300) = 22.09 kN-m
Muy1 = Muc + Mus(Total)
= 22.09 + (47.80)
= 69.90 kN-m
Pu/Puz = 0.793
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.793, an = 1.988
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((43.87/92.45)1.988) + ((44.81/69.90)1.988)
= 0.640 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 20 = 320 mm
               iii) 300 mm
               Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG29 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 14 nos. (4398 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG30 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 427.84 kN
MomentX,(Mx) = 3.73 kN-m
MomentY,(My) = 95.34 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C380.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 8.856 kN.m
My_MinEccen = 8.557 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 8.856 kN.m
My = max(My,My_MinEccen) + MuaddY = 95.337 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 3.73/427.84 = 9 mm
Actual eccenY = My / P = 95.34/427.84 = 223 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(9,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(223,20) = 223 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(223 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 16 nos. (1810 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00275 351.64 8.92 342.72 77.52 179 13.88
2 226 0.00192 324.96 8.91 316.05 71.49 128 9.14
3 226 0.00109 218.94 7.09 211.84 47.92 77 3.68
4 226 0.00027 53.02 2.21 50.81 11.49 26 0.29
5 226 -0.00056 -112.90 0.00 -112.90 -25.54 -26 0.65
6 226 -0.00139 -278.81 0.00 -278.81 -63.07 -77 4.84
7 226 -0.00222 -335.95 0.00 -335.95 -75.99 -128 9.72
8 226 -0.00305 -354.37 0.00 -354.37 -80.16 -179 14.35
Total-36.33 56.54
xux = 216 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 216/450 = 0.173
Puc = 0.173 x 20.00 x 300 x 450
= 466.07 kN
Pux1 = Puc + Pus(Total)
= 466.07 + (-36.33)
= 429.74 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 216/450
= 0.199
Muc = 466.07 x (0.5 x 450 - 0.199 x 450) = 63.03 kN-m
Mux1 = Muc + Mus(Total)
= 63.03 + (56.54)
= 119.57 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 905 0.00234 340.07 8.92 331.15 299.62 104 31.16
2 905 -0.00293 -353.28 0.00 -353.28 -319.64 -104 33.24
Total-20.02 64.40
xuy = 138 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 138/300 = 0.166
Puc = 0.166 x 20.00 x 450 x 300
= 448.03 kN
Puy1 = Puc + Pus(Total)
= 448.03 + (-20.02)
= 428.01 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 138/300
= 0.192
Muc = 448.03 x (0.5 x 300 - 0.192 x 300) = 41.43 kN-m
Muy1 = Muc + Mus(Total)
= 41.43 + (64.40)
= 105.83 kN-m
Pu/Puz = 0.243
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.243, an = 1.071
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((8.86/119.57)1.071) + ((95.34/105.83)1.071)
= 0.956 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG30 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 16 nos. (1810 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 873.52 kN
MomentX,(Mx) = 4.19 kN-m
MomentY,(My) = 86.65 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C380.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 18.082 kN.m
My_MinEccen = 17.470 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 18.082 kN.m
My = max(My,My_MinEccen) + MuaddY = 86.655 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 4.19/873.52 = 5 mm
Actual eccenY = My / P = 86.65/873.52 = 99 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(5,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(99,20) = 99 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(99 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 18 nos. (2036 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00296 353.53 8.92 344.61 77.95 179 13.95
2 226 0.00243 343.34 8.92 334.42 75.64 134 10.16
3 226 0.00190 323.77 8.90 314.88 71.22 90 6.37
4 226 0.00138 275.27 8.05 267.22 60.44 45 2.70
5 226 0.00085 169.82 5.97 163.86 37.06 0 0.00
6 226 0.00032 64.38 2.64 61.74 13.96 -45 -0.62
7 226 -0.00021 -41.07 0.00 -41.07 -9.29 -90 0.83
8 226 -0.00073 -146.52 0.00 -146.52 -33.14 -134 4.45
9 226 -0.00126 -251.96 0.00 -251.96 -56.99 -179 10.20
Total236.86 48.04
xux = 297 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 297/450 = 0.238
Puc = 0.238 x 20.00 x 300 x 450
= 641.67 kN
Pux1 = Puc + Pus(Total)
= 641.67 + (236.86)
= 878.53 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 297/450
= 0.275
Muc = 641.67 x (0.5 x 450 - 0.275 x 450) = 65.08 kN-m
Mux1 = Muc + Mus(Total)
= 65.08 + (48.04)
= 113.12 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1018 0.00273 351.09 8.92 342.17 348.29 104 36.22
2 1018 -0.00074 -147.61 0.00 -147.61 -150.25 -104 15.63
Total198.04 51.85
xuy = 210 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 210/300 = 0.252
Puc = 0.252 x 20.00 x 450 x 300
= 679.64 kN
Puy1 = Puc + Pus(Total)
= 679.64 + (198.04)
= 877.68 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 210/300
= 0.291
Muc = 679.64 x (0.5 x 300 - 0.291 x 300) = 42.64 kN-m
Muy1 = Muc + Mus(Total)
= 42.64 + (51.85)
= 94.49 kN-m
Pu/Puz = 0.477
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.477, an = 1.462
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((18.08/113.12)1.462) + ((86.65/94.49)1.462)
= 0.950 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG30 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 18 nos. (2036 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1315.99 kN
MomentX,(Mx) = 3.92 kN-m
MomentY,(My) = 83.11 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C380.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 27.241 kN.m
My_MinEccen = 26.320 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 27.241 kN.m
My = max(My,My_MinEccen) + MuaddY = 83.105 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 3.92/1315.99 = 3 mm
Actual eccenY = My / P = 83.11/1315.99 = 63 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(3,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(63,20) = 63 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(63 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 16 nos. (3217 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00301 354.01 8.92 345.09 138.77 177 24.56
2 402 0.00250 345.08 8.92 336.16 135.18 126 17.09
3 402 0.00198 327.18 8.92 318.26 127.98 76 9.71
4 402 0.00147 291.65 8.30 283.36 113.94 25 2.88
5 402 0.00096 191.48 6.50 184.99 74.39 -25 -1.88
6 402 0.00044 88.73 3.52 85.22 34.27 -76 -2.60
7 402 -0.00007 -14.01 0.00 -14.01 -5.64 -126 0.71
8 402 -0.00058 -116.76 0.00 -116.76 -46.95 -177 8.31
Total571.94 58.78
xux = 345 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 345/450 = 0.276
Puc = 0.276 x 20.00 x 300 x 450
= 744.19 kN
Pux1 = Puc + Pus(Total)
= 744.19 + (571.94)
= 1316.13 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 345/450
= 0.319
Muc = 744.19 x (0.5 x 450 - 0.319 x 450) = 60.78 kN-m
Mux1 = Muc + Mus(Total)
= 60.78 + (58.78)
= 119.57 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1608 0.00282 352.30 8.92 343.38 552.33 102 56.34
2 1608 -0.00008 -16.80 0.00 -16.80 -27.02 -102 2.76
Total525.31 59.09
xuy = 246 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 246/300 = 0.295
Puc = 0.295 x 20.00 x 450 x 300
= 797.34 kN
Puy1 = Puc + Pus(Total)
= 797.34 + (525.31)
= 1322.65 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 246/300
= 0.341
Muc = 797.34 x (0.5 x 300 - 0.341 x 300) = 37.97 kN-m
Muy1 = Muc + Mus(Total)
= 37.97 + (59.09)
= 97.07 kN-m
Pu/Puz = 0.602
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.602, an = 1.669
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((27.24/119.57)1.669) + ((83.11/97.07)1.669)
= 0.856 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG30 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 16 nos. (3217 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1753.42 kN
MomentX,(Mx) = 3.23 kN-m
MomentY,(My) = 77.95 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C380.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 36.296 kN.m
My_MinEccen = 35.068 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 36.296 kN.m
My = max(My,My_MinEccen) + MuaddY = 77.952 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 3.23/1753.42 = 2 mm
Actual eccenY = My / P = 77.95/1753.42 = 44 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(2,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(44,20) = 44 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(44 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 14 nos. (4398 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00304 354.28 8.92 345.36 217.00 175 37.97
2 628 0.00251 345.39 8.92 336.47 211.41 117 24.66
3 628 0.00198 326.92 8.92 318.00 199.81 58 11.66
4 628 0.00144 289.17 8.23 280.94 176.52 0 0.00
5 628 0.00091 182.43 6.28 176.15 110.68 -58 -6.46
6 628 0.00038 75.87 3.06 72.81 45.75 -117 -5.34
7 628 -0.00015 -30.68 0.00 -30.68 -19.28 -175 3.37
Total941.88 65.87
xux = 383 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 383/450 = 0.307
Puc = 0.307 x 20.00 x 300 x 450
= 827.72 kN
Pux1 = Puc + Pus(Total)
= 827.72 + (941.88)
= 1769.60 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 383/450
= 0.354
Muc = 827.72 x (0.5 x 450 - 0.354 x 450) = 54.29 kN-m
Mux1 = Muc + Mus(Total)
= 54.29 + (65.87)
= 120.16 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 2199 0.00286 352.64 8.92 343.72 755.89 100 75.59
2 2199 0.00028 56.32 2.34 53.99 118.72 -100 -11.87
Total874.61 63.72
xuy = 272 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 272/300 = 0.326
Puc = 0.326 x 20.00 x 450 x 300
= 880.88 kN
Puy1 = Puc + Pus(Total)
= 880.88 + (874.61)
= 1755.48 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 272/300
= 0.377
Muc = 880.88 x (0.5 x 300 - 0.377 x 300) = 32.50 kN-m
Muy1 = Muc + Mus(Total)
= 32.50 + (63.72)
= 96.22 kN-m
Pu/Puz = 0.689
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.689, an = 1.815
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((36.30/120.16)1.815) + ((77.95/96.22)1.815)
= 0.796 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 20 = 320 mm
               iii) 300 mm
               Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG30 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 14 nos. (4398 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 2179.33 kN
MomentX,(Mx) = 2.05 kN-m
MomentY,(My) = 50.30 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C380.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 47.400 kN.m
My_MinEccen = 43.587 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 47.400 kN.m
My = max(My,My_MinEccen) + MuaddY = 50.297 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 2.05/2179.33 = 1 mm
Actual eccenY = My / P = 50.30/2179.33 = 23 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(1,22) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(23,20) = 23 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(23 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 16 nos. (5027 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00310 354.73 8.92 345.81 217.28 175 38.02
2 628 0.00269 350.02 8.92 341.10 214.32 125 26.79
3 628 0.00229 338.24 8.92 329.32 206.92 75 15.52
4 628 0.00188 322.38 8.89 313.49 196.97 25 4.92
5 628 0.00148 292.16 8.31 283.85 178.35 -25 -4.46
6 628 0.00107 214.36 7.00 207.36 130.29 -75 -9.77
7 628 0.00067 133.42 4.96 128.47 80.72 -125 -10.09
8 628 0.00026 52.48 2.19 50.30 31.60 -175 -5.53
Total1256.45 55.41
xux = 432 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 432/450 = 0.346
Puc = 0.346 x 20.00 x 300 x 450
= 934.03 kN
Pux1 = Puc + Pus(Total)
= 934.03 + (1256.45)
= 2190.48 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 432/450
= 0.400
Muc = 934.03 x (0.5 x 450 - 0.400 x 450) = 42.14 kN-m
Mux1 = Muc + Mus(Total)
= 42.14 + (55.41)
= 97.54 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 2513 0.00287 352.75 8.92 343.83 864.15 100 86.42
2 2513 0.00066 131.36 4.90 126.46 317.83 -100 -31.78
Total1181.98 54.63
xuy = 309 mm               Puc = C1.fck.B.D
ku = 1.031
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.027
C1 = 0.446 x (1 - C3/6) = 0.370
Puc = 0.370 x 20.00 x 450 x 300
= 998.00 kN
Puy1 = Puc + Pus(Total)
= 998.00 + (1181.98)
= 2179.98 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.426
Muc = 998.00 x (0.5 x 300 - 0.426 x 300) = 22.09 kN-m
Muy1 = Muc + Mus(Total)
= 22.09 + (54.63)
= 76.72 kN-m
Pu/Puz = 0.797
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.797, an = 1.995
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((47.40/97.54)1.995) + ((50.30/76.72)1.995)
= 0.668 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 20 = 320 mm
               iii) 300 mm
               Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG30 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 16 nos. (5027 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG31 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 666.76 kN
MomentX,(Mx) = 53.85 kN-m
MomentY,(My) = 42.88 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C390.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 13.802 kN.m
My_MinEccen = 13.335 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 53.845 kN.m
My = max(My,My_MinEccen) + MuaddY = 42.877 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 53.85/666.76 = 81 mm
Actual eccenY = My / P = 42.88/666.76 = 64 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(81,21) = 81 mm
eccenY = max(Actual eccenY,eccenYMin) = max(64,20) = 64 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(81 mm) > 0.05 x 300(15 mm)
and eccenY(64 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00291 353.10 8.92 344.18 77.85 179 13.94
2 226 0.00176 314.78 8.79 305.98 69.21 90 6.19
3 226 0.00061 121.94 4.61 117.33 26.54 0 0.00
4 226 -0.00054 -108.01 0.00 -108.01 -24.43 -90 2.19
5 226 -0.00169 -310.43 0.00 -310.43 -70.22 -179 12.57
Total78.96 34.89
xux = 272 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 272/450 = 0.218
Puc = 0.218 x 20.00 x 300 x 450
= 588.52 kN
Pux1 = Puc + Pus(Total)
= 588.52 + (78.96)
= 667.47 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 272/450
= 0.252
Muc = 588.52 x (0.5 x 450 - 0.252 x 450) = 65.71 kN-m
Mux1 = Muc + Mus(Total)
= 65.71 + (34.89)
= 100.60 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00265 348.95 8.92 340.03 192.28 104 20.00
2 565 -0.00120 -239.46 0.00 -239.46 -135.41 -104 14.08
Total56.87 34.08
xuy = 189 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 189/300 = 0.227
Puc = 0.227 x 20.00 x 450 x 300
= 613.20 kN
Puy1 = Puc + Pus(Total)
= 613.20 + (56.87)
= 670.07 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 189/300
= 0.262
Muc = 613.20 x (0.5 x 300 - 0.262 x 300) = 43.70 kN-m
Muy1 = Muc + Mus(Total)
= 43.70 + (34.08)
= 77.78 kN-m
Pu/Puz = 0.428
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.428, an = 1.380
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((53.85/100.60)1.380) + ((42.88/77.78)1.380)
= 0.861 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG31 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1313.45 kN
MomentX,(Mx) = 46.28 kN-m
MomentY,(My) = 40.29 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C390.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 27.188 kN.m
My_MinEccen = 26.269 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 46.277 kN.m
My = max(My,My_MinEccen) + MuaddY = 40.294 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 46.28/1313.45 = 35 mm
Actual eccenY = My / P = 40.29/1313.45 = 31 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(35,21) = 35 mm
eccenY = max(Actual eccenY,eccenYMin) = max(31,20) = 31 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(35 mm) > 0.05 x 300(15 mm)
and eccenY(31 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 2 nos. + #16 - 8 nos. (1835 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00310 354.76 8.92 345.84 139.07 177 24.62
2 402 0.00235 340.58 8.92 331.66 133.37 88 11.67
3 226 0.00162 305.53 8.59 296.94 67.17 0 0.00
4 402 0.00089 177.12 6.15 170.97 68.75 -88 -6.02
5 402 0.00014 27.37 1.18 26.19 10.53 -177 -1.86
Total418.89 28.41
xux = 418 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 418/450 = 0.335
Puc = 0.335 x 20.00 x 300 x 450
= 903.66 kN
Pux1 = Puc + Pus(Total)
= 903.66 + (418.89)
= 1322.55 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 418/450
= 0.387
Muc = 903.66 x (0.5 x 450 - 0.387 x 450) = 46.05 kN-m
Mux1 = Muc + Mus(Total)
= 46.05 + (28.41)
= 74.46 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 917 0.00291 353.13 8.92 344.21 315.76 102 32.21
2 917 0.00042 83.08 3.32 79.76 73.17 -102 -7.46
Total388.93 24.74
xuy = 286 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 286/300 = 0.343
Puc = 0.343 x 20.00 x 450 x 300
= 926.44 kN
Puy1 = Puc + Pus(Total)
= 926.44 + (388.93)
= 1315.37 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 286/300
= 0.397
Muc = 926.44 x (0.5 x 300 - 0.397 x 300) = 28.77 kN-m
Muy1 = Muc + Mus(Total)
= 28.77 + (24.74)
= 53.51 kN-m
Pu/Puz = 0.742
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.742, an = 1.904
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((46.28/74.46)1.904) + ((40.29/53.51)1.904)
= 0.987 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG31 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 2 nos. + #16 - 8 nos. (1835 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1975.35 kN
MomentX,(Mx) = 37.10 kN-m
MomentY,(My) = 39.00 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C390.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 40.890 kN.m
My_MinEccen = 39.507 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 40.890 kN.m
My = max(My,My_MinEccen) + MuaddY = 39.507 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 37.10/1975.35 = 19 mm
Actual eccenY = My / P = 39.00/1975.35 = 20 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(19,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(20,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 12 nos. (3770 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00305 354.37 8.92 345.45 217.05 175 37.98
2 628 0.00254 346.07 8.92 337.15 211.84 105 22.24
3 628 0.00202 328.51 8.92 319.59 200.81 35 7.03
4 628 0.00150 294.84 8.37 286.47 180.00 -35 -6.30
5 628 0.00099 197.73 6.64 191.09 120.06 -105 -12.61
6 628 0.00047 94.49 3.72 90.77 57.03 -175 -9.98
Total986.79 38.37
xux = 464 mm               Puc = C1.fck.B.D
ku = 1.031
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.027
C1 = 0.446 x (1 - C3/6) = 0.370
Puc = 0.370 x 20.00 x 300 x 450
= 998.00 kN
Pux1 = Puc + Pus(Total)
= 998.00 + (986.79)
= 1984.79 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.426
Muc = 998.00 x (0.5 x 450 - 0.426 x 450) = 33.14 kN-m
Mux1 = Muc + Mus(Total)
= 33.14 + (38.37)
= 71.51 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1885 0.00277 351.88 8.92 342.96 646.47 100 64.65
2 1885 0.00081 162.19 5.77 156.42 294.84 -100 -29.48
Total941.31 35.16
xuy = 333 mm               Puc = C1.fck.B.D
ku = 1.109
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.805
C1 = 0.446 x (1 - C3/6) = 0.386
Puc = 0.386 x 20.00 x 450 x 300
= 1042.61 kN
Puy1 = Puc + Pus(Total)
= 1042.61 + (941.31)
= 1983.92 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.445
Muc = 1042.61 x (0.5 x 300 - 0.445 x 300) = 17.31 kN-m
Muy1 = Muc + Mus(Total)
= 17.31 + (35.16)
= 52.48 kN-m
Pu/Puz = 0.839
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.839, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((40.89/71.51)2.000) + ((39.51/52.48)2.000)
= 0.894 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 20 = 320 mm
               iii) 300 mm
               Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG31 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 12 nos. (3770 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 2661.52 kN
MomentX,(Mx) = 25.04 kN-m
MomentY,(My) = 36.47 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C390.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 55.094 kN.m
My_MinEccen = 53.230 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 55.094 kN.m
My = max(My,My_MinEccen) + MuaddY = 53.230 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 25.04/2661.52 = 9 mm
Actual eccenY = My / P = 36.47/2661.52 = 14 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(9,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(14,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #25 - 14 nos. (6872 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 982 0.00308 354.59 8.92 345.67 339.36 173 58.54
2 982 0.00262 348.12 8.92 339.20 333.01 115 38.30
3 982 0.00216 333.44 8.92 324.52 318.60 58 18.32
4 982 0.00169 310.67 8.71 301.96 296.44 0 0.00
5 982 0.00123 246.38 7.60 238.78 234.42 -58 -13.48
6 982 0.00077 154.05 5.55 148.50 145.79 -115 -16.77
7 982 0.00031 61.72 2.54 59.18 58.10 -173 -10.02
Total1725.72 74.89
xux = 436 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 436/450 = 0.349
Puc = 0.349 x 20.00 x 300 x 450
= 941.63 kN
Pux1 = Puc + Pus(Total)
= 941.63 + (1725.72)
= 2667.34 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 436/450
= 0.403
Muc = 941.63 x (0.5 x 450 - 0.403 x 450) = 41.10 kN-m
Mux1 = Muc + Mus(Total)
= 41.10 + (74.89)
= 115.99 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 3436 0.00282 352.33 8.92 343.41 1179.99 98 115.05
2 3436 0.00072 143.54 5.25 138.28 475.16 -98 -46.33
Total1655.15 68.72
xuy = 314 mm               Puc = C1.fck.B.D
ku = 1.047
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.976
C1 = 0.446 x (1 - C3/6) = 0.373
Puc = 0.373 x 20.00 x 450 x 300
= 1008.29 kN
Puy1 = Puc + Pus(Total)
= 1008.29 + (1655.15)
= 2663.44 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.431
Muc = 1008.29 x (0.5 x 300 - 0.431 x 300) = 20.99 kN-m
Muy1 = Muc + Mus(Total)
= 20.99 + (68.72)
= 89.71 kN-m
Pu/Puz = 0.808
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.808, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((55.09/115.99)2.000) + ((53.23/89.71)2.000)
= 0.578 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 25/4 = 6 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (6, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 25 = 400 mm
               iii) 300 mm
               Required spacing = minimum of (300, 400, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG31 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #25 - 14 nos. (6872 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 3403.81 kN
MomentX,(Mx) = 10.14 kN-m
MomentY,(My) = 22.72 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C390.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 74.033 kN.m
My_MinEccen = 68.076 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 74.033 kN.m
My = max(My,My_MinEccen) + MuaddY = 68.076 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 10.14/3403.81 = 3 mm
Actual eccenY = My / P = 22.72/3403.81 = 7 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(3,22) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(7,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #32 - 12 nos. (9651 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1608 0.00306 354.40 8.92 345.48 555.71 169 93.91
2 1608 0.00252 345.72 8.92 336.80 541.74 101 54.93
3 1608 0.00199 327.35 8.92 318.43 512.19 34 17.31
4 1608 0.00146 290.14 8.26 281.88 453.40 -34 -15.32
5 1608 0.00092 184.21 6.32 177.88 286.12 -101 -29.01
6 1608 0.00039 77.38 3.12 74.27 119.46 -169 -20.19
Total2468.61 101.63
xux = 443 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 443/450 = 0.354
Puc = 0.354 x 20.00 x 300 x 450
= 956.81 kN
Pux1 = Puc + Pus(Total)
= 956.81 + (2468.61)
= 3425.42 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 443/450
= 0.410
Muc = 956.81 x (0.5 x 450 - 0.410 x 450) = 38.97 kN-m
Mux1 = Muc + Mus(Total)
= 38.97 + (101.63)
= 140.60 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 4825 0.00276 351.83 8.92 342.91 1654.70 94 155.54
2 4825 0.00079 157.22 5.63 151.59 731.48 -94 -68.76
Total2386.18 86.78
xuy = 319 mm               Puc = C1.fck.B.D
ku = 1.063
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.929
C1 = 0.446 x (1 - C3/6) = 0.377
Puc = 0.377 x 20.00 x 450 x 300
= 1017.83 kN
Puy1 = Puc + Pus(Total)
= 1017.83 + (2386.18)
= 3404.00 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.435
Muc = 1017.83 x (0.5 x 300 - 0.435 x 300) = 19.97 kN-m
Muy1 = Muc + Mus(Total)
= 19.97 + (86.78)
= 106.75 kN-m
Pu/Puz = 0.824
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.824, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((74.03/140.60)2.000) + ((68.08/106.75)2.000)
= 0.684 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 32/4 = 8 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (8, 6) = 8 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 32 = 512 mm
               iii) 300 mm
               Required spacing = minimum of (300, 512, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG31 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #32 - 12 nos. (9651 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG32 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 599.40 kN
MomentX,(Mx) = 93.71 kN-m
MomentY,(My) = 5.22 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C400.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 12.407 kN.m
My_MinEccen = 11.988 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 93.705 kN.m
My = max(My,My_MinEccen) + MuaddY = 11.988 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 93.71/599.40 = 156 mm
Actual eccenY = My / P = 5.22/599.40 = 9 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(156,21) = 156 mm
eccenY = max(Actual eccenY,eccenYMin) = max(9,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(156 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00287 352.75 8.92 343.83 77.77 179 13.92
2 226 0.00164 307.28 8.63 298.65 67.55 90 6.05
3 226 0.00041 82.07 3.28 78.78 17.82 0 0.00
4 226 -0.00082 -163.73 0.00 -163.73 -37.03 -90 3.31
5 226 -0.00205 -329.49 0.00 -329.49 -74.53 -179 13.34
Total51.58 36.62
xux = 255 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 255/450 = 0.204
Puc = 0.204 x 20.00 x 300 x 450
= 550.55 kN
Pux1 = Puc + Pus(Total)
= 550.55 + (51.58)
= 602.13 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 255/450
= 0.236
Muc = 550.55 x (0.5 x 450 - 0.236 x 450) = 65.50 kN-m
Mux1 = Muc + Mus(Total)
= 65.50 + (36.62)
= 102.12 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00260 347.59 8.92 338.67 191.51 104 19.92
2 565 -0.00149 -293.52 0.00 -293.52 -165.98 -104 17.26
Total25.53 37.18
xuy = 178 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 178/300 = 0.214
Puc = 0.214 x 20.00 x 450 x 300
= 577.13 kN
Puy1 = Puc + Pus(Total)
= 577.13 + (25.53)
= 602.65 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 178/300
= 0.247
Muc = 577.13 x (0.5 x 300 - 0.247 x 300) = 43.80 kN-m
Muy1 = Muc + Mus(Total)
= 43.80 + (37.18)
= 80.98 kN-m
Pu/Puz = 0.385
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.385, an = 1.308
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((93.71/102.12)1.308) + ((11.99/80.98)1.308)
= 0.976 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG32 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1216.59 kN
MomentX,(Mx) = 85.57 kN-m
MomentY,(My) = 4.50 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C400.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 25.184 kN.m
My_MinEccen = 24.332 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 85.567 kN.m
My = max(My,My_MinEccen) + MuaddY = 24.332 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 85.57/1216.59 = 70 mm
Actual eccenY = My / P = 4.50/1216.59 = 4 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(70,21) = 70 mm
eccenY = max(Actual eccenY,eccenYMin) = max(4,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(70 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 8 nos. + #16 - 6 nos. (2111 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00305 354.33 8.92 345.41 138.90 177 24.58
2 226 0.00249 344.83 8.92 335.91 75.98 117 8.92
3 226 0.00195 325.80 8.91 316.88 71.68 60 4.28
4 402 0.00139 277.36 8.08 269.28 108.28 -0 -0.00
5 226 0.00083 165.28 5.85 159.43 36.06 -60 -2.15
6 226 0.00028 56.96 2.36 54.60 12.35 -117 -1.45
7 402 -0.00028 -55.12 0.00 -55.12 -22.16 -177 3.92
Total421.09 38.10
xux = 373 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 373/450 = 0.298
Puc = 0.298 x 20.00 x 300 x 450
= 804.94 kN
Pux1 = Puc + Pus(Total)
= 804.94 + (421.09)
= 1226.02 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 373/450
= 0.345
Muc = 804.94 x (0.5 x 450 - 0.345 x 450) = 56.33 kN-m
Mux1 = Muc + Mus(Total)
= 56.33 + (38.10)
= 94.42 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1056 0.00285 352.62 8.92 343.70 362.81 102 37.01
2 1056 0.00011 21.95 0.95 20.99 22.16 -102 -2.26
Total384.97 34.75
xuy = 260 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 260/300 = 0.312
Puc = 0.312 x 20.00 x 450 x 300
= 842.91 kN
Puy1 = Puc + Pus(Total)
= 842.91 + (384.97)
= 1227.87 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 260/300
= 0.361
Muc = 842.91 x (0.5 x 300 - 0.361 x 300) = 35.21 kN-m
Muy1 = Muc + Mus(Total)
= 35.21 + (34.75)
= 69.96 kN-m
Pu/Puz = 0.657
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.657, an = 1.761
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((85.57/94.42)1.761) + ((24.33/69.96)1.761)
= 0.996 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG32 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 8 nos. + #16 - 6 nos. (2111 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1807.15 kN
MomentX,(Mx) = 74.81 kN-m
MomentY,(My) = 2.90 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C400.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 37.408 kN.m
My_MinEccen = 36.143 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 74.805 kN.m
My = max(My,My_MinEccen) + MuaddY = 36.143 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 74.81/1807.15 = 41 mm
Actual eccenY = My / P = 2.90/1807.15 = 2 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(41,21) = 41 mm
eccenY = max(Actual eccenY,eccenYMin) = max(2,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(41 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 12 nos. (3770 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00309 354.65 8.92 345.73 217.23 175 38.01
2 628 0.00250 345.23 8.92 336.31 211.31 105 22.19
3 628 0.00192 324.94 8.91 316.03 198.57 35 6.95
4 628 0.00134 268.59 7.96 260.64 163.76 -35 -5.73
5 628 0.00076 152.44 5.50 146.94 92.33 -105 -9.69
6 628 0.00018 36.30 1.55 34.75 21.83 -175 -3.82
Total905.02 47.90
xux = 422 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 422/450 = 0.338
Puc = 0.338 x 20.00 x 300 x 450
= 911.25 kN
Pux1 = Puc + Pus(Total)
= 911.25 + (905.02)
= 1816.27 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 422/450
= 0.390
Muc = 911.25 x (0.5 x 450 - 0.390 x 450) = 45.11 kN-m
Mux1 = Muc + Mus(Total)
= 45.11 + (47.90)
= 93.01 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1885 0.00291 353.13 8.92 344.21 648.82 100 64.88
2 1885 0.00056 112.07 4.30 107.78 203.15 -100 -20.32
Total851.97 44.57
xuy = 298 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 298/300 = 0.357
Puc = 0.357 x 20.00 x 450 x 300
= 964.41 kN
Puy1 = Puc + Pus(Total)
= 964.41 + (851.97)
= 1816.38 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 298/300
= 0.413
Muc = 964.41 x (0.5 x 300 - 0.413 x 300) = 25.24 kN-m
Muy1 = Muc + Mus(Total)
= 25.24 + (44.57)
= 69.81 kN-m
Pu/Puz = 0.768
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.768, an = 1.946
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((74.81/93.01)1.946) + ((36.14/69.81)1.946)
= 0.932 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 20 = 320 mm
               iii) 300 mm
               Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG32 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 12 nos. (3770 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 2350.21 kN
MomentX,(Mx) = 60.04 kN-m
MomentY,(My) = 0.99 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C400.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 48.649 kN.m
My_MinEccen = 47.004 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 60.042 kN.m
My = max(My,My_MinEccen) + MuaddY = 47.004 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 60.04/2350.21 = 26 mm
Actual eccenY = My / P = 0.99/2350.21 = 0 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(26,21) = 26 mm
eccenY = max(Actual eccenY,eccenYMin) = max(0,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(26 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #25 - 12 nos. (5890 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 982 0.00307 354.50 8.92 345.58 339.27 173 58.52
2 982 0.00250 345.12 8.92 336.20 330.07 104 34.16
3 982 0.00193 325.26 8.91 316.35 310.58 35 10.71
4 982 0.00136 272.98 8.02 264.96 260.12 -35 -8.97
5 982 0.00080 159.44 5.69 153.74 150.94 -104 -15.62
6 982 0.00023 45.90 1.93 43.97 43.16 -173 -7.45
Total1434.14 71.36
xux = 425 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 425/450 = 0.340
Puc = 0.340 x 20.00 x 300 x 450
= 918.84 kN
Pux1 = Puc + Pus(Total)
= 918.84 + (1434.14)
= 2352.98 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 425/450
= 0.393
Muc = 918.84 x (0.5 x 450 - 0.393 x 450) = 44.14 kN-m
Mux1 = Muc + Mus(Total)
= 44.14 + (71.36)
= 115.50 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 2945 0.00286 352.71 8.92 343.79 1012.54 98 98.72
2 2945 0.00065 129.89 4.85 125.03 368.25 -98 -35.90
Total1380.80 62.82
xuy = 305 mm               Puc = C1.fck.B.D
ku = 1.016
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.083
C1 = 0.446 x (1 - C3/6) = 0.366
Puc = 0.366 x 20.00 x 450 x 300
= 986.88 kN
Puy1 = Puc + Pus(Total)
= 986.88 + (1380.80)
= 2367.67 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.421
Muc = 986.88 x (0.5 x 300 - 0.421 x 300) = 23.28 kN-m
Muy1 = Muc + Mus(Total)
= 23.28 + (62.82)
= 86.10 kN-m
Pu/Puz = 0.785
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.785, an = 1.974
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((60.04/115.50)1.974) + ((47.00/86.10)1.974)
= 0.578 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 25/4 = 6 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (6, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 25 = 400 mm
               iii) 300 mm
               Required spacing = minimum of (300, 400, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG32 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #25 - 12 nos. (5890 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 2789.30 kN
MomentX,(Mx) = 34.21 kN-m
MomentY,(My) = 1.48 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C400.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 60.667 kN.m
My_MinEccen = 55.786 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 60.667 kN.m
My = max(My,My_MinEccen) + MuaddY = 55.786 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 34.21/2789.30 = 12 mm
Actual eccenY = My / P = 1.48/2789.30 = 1 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(12,22) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(1,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #25 - 14 nos. (6872 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 982 0.00304 354.21 8.92 345.29 338.98 173 58.47
2 982 0.00261 347.97 8.92 339.05 332.86 115 38.28
3 982 0.00219 334.61 8.92 325.69 319.74 58 18.39
4 982 0.00176 315.00 8.79 306.20 300.62 0 0.00
5 982 0.00134 267.79 7.95 259.84 255.10 -58 -14.67
6 982 0.00091 182.98 6.29 176.69 173.46 -115 -19.95
7 982 0.00049 98.17 3.84 94.33 92.61 -173 -15.98
Total1813.37 64.55
xux = 464 mm               Puc = C1.fck.B.D
ku = 1.031
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.027
C1 = 0.446 x (1 - C3/6) = 0.370
Puc = 0.370 x 20.00 x 300 x 450
= 998.00 kN
Pux1 = Puc + Pus(Total)
= 998.00 + (1813.37)
= 2811.37 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.426
Muc = 998.00 x (0.5 x 450 - 0.426 x 450) = 33.14 kN-m
Mux1 = Muc + Mus(Total)
= 33.14 + (64.55)
= 97.69 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 3436 0.00273 350.98 8.92 342.06 1175.37 98 114.60
2 3436 0.00086 172.31 6.03 166.28 571.35 -98 -55.71
Total1746.72 58.89
xuy = 338 mm               Puc = C1.fck.B.D
ku = 1.125
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.769
C1 = 0.446 x (1 - C3/6) = 0.389
Puc = 0.389 x 20.00 x 450 x 300
= 1049.78 kN
Puy1 = Puc + Pus(Total)
= 1049.78 + (1746.72)
= 2796.49 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.447
Muc = 1049.78 x (0.5 x 300 - 0.447 x 300) = 16.55 kN-m
Muy1 = Muc + Mus(Total)
= 16.55 + (58.89)
= 75.44 kN-m
Pu/Puz = 0.847
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.847, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((60.67/97.69)2.000) + ((55.79/75.44)2.000)
= 0.933 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 25/4 = 6 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (6, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 25 = 400 mm
               iii) 300 mm
               Required spacing = minimum of (300, 400, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG32 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #25 - 14 nos. (6872 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG33 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 580.95 kN
MomentX,(Mx) = 122.66 kN-m
MomentY,(My) = 7.38 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C410.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 12.026 kN.m
My_MinEccen = 11.619 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 122.662 kN.m
My = max(My,My_MinEccen) + MuaddY = 11.619 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 122.66/580.95 = 211 mm
Actual eccenY = My / P = 7.38/580.95 = 13 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(211,21) = 211 mm
eccenY = max(Actual eccenY,eccenYMin) = max(13,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(211 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 2 nos. + #16 - 8 nos. (1835 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00281 352.28 8.92 343.36 138.07 177 24.44
2 402 0.00154 297.93 8.44 289.49 116.41 88 10.19
3 226 0.00029 57.71 2.39 55.32 12.51 0 0.00
4 402 -0.00096 -192.07 0.00 -192.07 -77.24 -88 6.76
5 402 -0.00224 -336.48 0.00 -336.48 -135.30 -177 23.95
Total54.45 65.33
xux = 245 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 245/450 = 0.196
Puc = 0.196 x 20.00 x 300 x 450
= 529.66 kN
Pux1 = Puc + Pus(Total)
= 529.66 + (54.45)
= 584.12 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 245/450
= 0.227
Muc = 529.66 x (0.5 x 450 - 0.227 x 450) = 65.14 kN-m
Mux1 = Muc + Mus(Total)
= 65.14 + (65.33)
= 130.48 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 917 0.00252 345.75 8.92 336.83 308.99 102 31.52
2 917 -0.00162 -305.75 0.00 -305.75 -280.48 -102 28.61
Total28.51 60.13
xuy = 172 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 172/300 = 0.207
Puc = 0.207 x 20.00 x 450 x 300
= 558.14 kN
Puy1 = Puc + Pus(Total)
= 558.14 + (28.51)
= 586.65 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 172/300
= 0.239
Muc = 558.14 x (0.5 x 300 - 0.239 x 300) = 43.72 kN-m
Muy1 = Muc + Mus(Total)
= 43.72 + (60.13)
= 103.85 kN-m
Pu/Puz = 0.328
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.328, an = 1.214
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((122.66/130.48)1.214) + ((11.62/103.85)1.214)
= 0.998 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG33 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 2 nos. + #16 - 8 nos. (1835 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1167.62 kN
MomentX,(Mx) = 114.49 kN-m
MomentY,(My) = 6.90 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C410.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 24.170 kN.m
My_MinEccen = 23.352 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 114.488 kN.m
My = max(My,My_MinEccen) + MuaddY = 23.352 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 114.49/1167.62 = 98 mm
Actual eccenY = My / P = 6.90/1167.62 = 6 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(98,21) = 98 mm
eccenY = max(Actual eccenY,eccenYMin) = max(6,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(98 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 16 nos. (3217 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00298 353.71 8.92 344.79 138.65 177 24.54
2 402 0.00243 343.25 8.92 334.33 134.44 126 17.00
3 402 0.00188 322.13 8.89 313.25 125.96 76 9.56
4 402 0.00133 265.41 7.91 257.50 103.55 25 2.62
5 402 0.00078 155.36 5.58 149.78 60.23 -25 -1.52
6 402 0.00023 45.31 1.91 43.41 17.45 -76 -1.32
7 402 -0.00032 -64.74 0.00 -64.74 -26.03 -126 3.29
8 402 -0.00087 -174.78 0.00 -174.78 -70.28 -177 12.44
Total483.96 66.60
xux = 322 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 322/450 = 0.257
Puc = 0.257 x 20.00 x 300 x 450
= 694.83 kN
Pux1 = Puc + Pus(Total)
= 694.83 + (483.96)
= 1178.79 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 322/450
= 0.297
Muc = 694.83 x (0.5 x 450 - 0.297 x 450) = 63.36 kN-m
Mux1 = Muc + Mus(Total)
= 63.36 + (66.60)
= 129.95 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1608 0.00276 351.84 8.92 342.92 551.59 102 56.26
2 1608 -0.00036 -71.94 0.00 -71.94 -115.71 -102 11.80
Total435.88 68.06
xuy = 229 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 229/300 = 0.274
Puc = 0.274 x 20.00 x 450 x 300
= 740.39 kN
Puy1 = Puc + Pus(Total)
= 740.39 + (435.88)
= 1176.27 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 229/300
= 0.317
Muc = 740.39 x (0.5 x 300 - 0.317 x 300) = 40.68 kN-m
Muy1 = Muc + Mus(Total)
= 40.68 + (68.06)
= 108.74 kN-m
Pu/Puz = 0.534
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.534, an = 1.556
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((114.49/129.95)1.556) + ((23.35/108.74)1.556)
= 0.912 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG33 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 16 nos. (3217 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1774.51 kN
MomentX,(Mx) = 104.45 kN-m
MomentY,(My) = 6.07 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C410.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 36.732 kN.m
My_MinEccen = 35.490 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 104.449 kN.m
My = max(My,My_MinEccen) + MuaddY = 35.490 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 104.45/1774.51 = 59 mm
Actual eccenY = My / P = 6.07/1774.51 = 3 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(59,21) = 59 mm
eccenY = max(Actual eccenY,eccenYMin) = max(3,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(59 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 14 nos. (4398 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00305 354.32 8.92 345.40 217.02 175 37.98
2 628 0.00252 345.62 8.92 336.70 211.55 117 24.68
3 628 0.00199 327.43 8.92 318.51 200.13 58 11.67
4 628 0.00146 290.94 8.28 282.66 177.60 0 0.00
5 628 0.00094 187.14 6.39 180.74 113.56 -58 -6.62
6 628 0.00041 81.55 3.27 78.28 49.19 -117 -5.74
7 628 -0.00012 -24.04 0.00 -24.04 -15.11 -175 2.64
Total953.94 64.61
xux = 387 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 387/450 = 0.309
Puc = 0.309 x 20.00 x 300 x 450
= 835.31 kN
Pux1 = Puc + Pus(Total)
= 835.31 + (953.94)
= 1789.26 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 387/450
= 0.358
Muc = 835.31 x (0.5 x 450 - 0.358 x 450) = 53.56 kN-m
Mux1 = Muc + Mus(Total)
= 53.56 + (64.61)
= 118.18 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 2199 0.00286 352.69 8.92 343.77 755.99 100 75.60
2 2199 0.00031 61.82 2.54 59.28 130.36 -100 -13.04
Total886.35 62.56
xuy = 274 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 274/300 = 0.329
Puc = 0.329 x 20.00 x 450 x 300
= 888.47 kN
Puy1 = Puc + Pus(Total)
= 888.47 + (886.35)
= 1774.82 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 274/300
= 0.380
Muc = 888.47 x (0.5 x 300 - 0.380 x 300) = 31.92 kN-m
Muy1 = Muc + Mus(Total)
= 31.92 + (62.56)
= 94.48 kN-m
Pu/Puz = 0.697
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.697, an = 1.829
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((104.45/118.18)1.829) + ((35.49/94.48)1.829)
= 0.965 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 20 = 320 mm
               iii) 300 mm
               Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG33 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 14 nos. (4398 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 2420.73 kN
MomentX,(Mx) = 92.18 kN-m
MomentY,(My) = 4.69 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C410.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 50.109 kN.m
My_MinEccen = 48.415 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 92.184 kN.m
My = max(My,My_MinEccen) + MuaddY = 48.415 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 92.18/2420.73 = 38 mm
Actual eccenY = My / P = 4.69/2420.73 = 2 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(38,21) = 38 mm
eccenY = max(Actual eccenY,eccenYMin) = max(2,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(38 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #25 - 14 nos. (6872 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 982 0.00304 354.26 8.92 345.34 339.04 173 58.48
2 982 0.00254 346.13 8.92 337.21 331.05 115 38.07
3 982 0.00204 329.11 8.92 320.19 314.34 58 18.07
4 982 0.00154 297.71 8.44 289.27 283.99 0 0.00
5 982 0.00103 206.59 6.83 199.75 196.11 -58 -11.28
6 982 0.00053 106.16 4.11 102.05 100.19 -115 -11.52
7 982 0.00003 5.73 0.25 5.48 5.38 -173 -0.93
Total1570.10 90.90
xux = 401 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 401/450 = 0.321
Puc = 0.321 x 20.00 x 300 x 450
= 865.69 kN
Pux1 = Puc + Pus(Total)
= 865.69 + (1570.10)
= 2435.79 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 401/450
= 0.370
Muc = 865.69 x (0.5 x 450 - 0.370 x 450) = 50.45 kN-m
Mux1 = Muc + Mus(Total)
= 50.45 + (90.90)
= 141.35 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 3436 0.00286 352.70 8.92 343.78 1181.26 98 115.17
2 3436 0.00050 99.02 3.87 95.15 326.96 -98 -31.88
Total1508.22 83.29
xuy = 288 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 288/300 = 0.346
Puc = 0.346 x 20.00 x 450 x 300
= 934.03 kN
Puy1 = Puc + Pus(Total)
= 934.03 + (1508.22)
= 2442.25 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 288/300
= 0.400
Muc = 934.03 x (0.5 x 300 - 0.400 x 300) = 28.09 kN-m
Muy1 = Muc + Mus(Total)
= 28.09 + (83.29)
= 111.39 kN-m
Pu/Puz = 0.735
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.735, an = 1.892
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((92.18/141.35)1.892) + ((48.41/111.39)1.892)
= 0.652 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 25/4 = 6 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (6, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 25 = 400 mm
               iii) 300 mm
               Required spacing = minimum of (300, 400, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG33 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #25 - 14 nos. (6872 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 3144.61 kN
MomentX,(Mx) = 60.67 kN-m
MomentY,(My) = 2.50 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C410.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 68.395 kN.m
My_MinEccen = 62.892 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 68.395 kN.m
My = max(My,My_MinEccen) + MuaddY = 62.892 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 60.67/3144.61 = 19 mm
Actual eccenY = My / P = 2.50/3144.61 = 1 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(19,22) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(1,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #32 - 12 nos. (9651 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1608 0.00302 354.11 8.92 345.19 555.23 169 93.83
2 1608 0.00245 343.78 8.92 334.86 538.63 101 54.62
3 1608 0.00187 321.87 8.88 312.99 503.44 34 17.02
4 1608 0.00130 259.57 7.82 251.75 404.94 -34 -13.69
5 1608 0.00072 144.53 5.28 139.25 223.98 -101 -22.71
6 1608 0.00015 29.49 1.27 28.22 45.40 -169 -7.67
Total2271.62 121.40
xux = 411 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 411/450 = 0.329
Puc = 0.329 x 20.00 x 300 x 450
= 888.47 kN
Pux1 = Puc + Pus(Total)
= 888.47 + (2271.62)
= 3160.08 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 411/450
= 0.380
Muc = 888.47 x (0.5 x 450 - 0.380 x 450) = 47.88 kN-m
Mux1 = Muc + Mus(Total)
= 47.88 + (121.40)
= 169.27 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 4825 0.00283 352.42 8.92 343.50 1657.56 94 155.81
2 4825 0.00059 117.00 4.46 112.55 543.10 -94 -51.05
Total2200.66 104.76
xuy = 293 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 293/300 = 0.352
Puc = 0.352 x 20.00 x 450 x 300
= 949.22 kN
Puy1 = Puc + Pus(Total)
= 949.22 + (2200.66)
= 3149.88 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 293/300
= 0.406
Muc = 949.22 x (0.5 x 300 - 0.406 x 300) = 26.70 kN-m
Muy1 = Muc + Mus(Total)
= 26.70 + (104.76)
= 131.46 kN-m
Pu/Puz = 0.761
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.761, an = 1.935
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((68.40/169.27)1.935) + ((62.89/131.46)1.935)
= 0.413 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 32/4 = 8 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (8, 6) = 8 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 32 = 512 mm
               iii) 300 mm
               Required spacing = minimum of (300, 512, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG33 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #32 - 12 nos. (9651 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG34 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 511.88 kN
MomentX,(Mx) = 40.16 kN-m
MomentY,(My) = 1.19 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C420.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 10.596 kN.m
My_MinEccen = 10.238 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 40.160 kN.m
My = max(My,My_MinEccen) + MuaddY = 10.238 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 40.16/511.88 = 78 mm
Actual eccenY = My / P = 1.19/511.88 = 2 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(78,21) = 78 mm
eccenY = max(Actual eccenY,eccenYMin) = max(2,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(78 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00281 352.20 8.92 343.28 77.65 179 13.90
2 226 0.00146 290.22 8.26 281.96 63.78 90 5.71
3 226 0.00011 21.21 0.92 20.29 4.59 0 0.00
4 226 -0.00124 -248.79 0.00 -248.79 -56.28 -90 5.04
5 226 -0.00259 -347.53 0.00 -347.53 -78.61 -179 14.07
Total11.13 38.72
xux = 232 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 232/450 = 0.186
Puc = 0.186 x 20.00 x 300 x 450
= 501.19 kN
Pux1 = Puc + Pus(Total)
= 501.19 + (11.13)
= 512.32 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 232/450
= 0.215
Muc = 501.19 x (0.5 x 450 - 0.215 x 450) = 64.39 kN-m
Mux1 = Muc + Mus(Total)
= 64.39 + (38.72)
= 103.11 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00248 344.66 8.92 335.74 189.86 104 19.75
2 565 -0.00212 -332.12 0.00 -332.12 -187.81 -104 19.53
Total2.04 39.28
xuy = 158 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 158/300 = 0.190
Puc = 0.190 x 20.00 x 450 x 300
= 512.58 kN
Puy1 = Puc + Pus(Total)
= 512.58 + (2.04)
= 514.62 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 158/300
= 0.219
Muc = 512.58 x (0.5 x 300 - 0.219 x 300) = 43.15 kN-m
Muy1 = Muc + Mus(Total)
= 43.15 + (39.28)
= 82.43 kN-m
Pu/Puz = 0.329
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.329, an = 1.215
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((40.16/103.11)1.215) + ((10.24/82.43)1.215)
= 0.398 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG34 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1050.90 kN
MomentX,(Mx) = 33.14 kN-m
MomentY,(My) = 0.58 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C420.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 21.754 kN.m
My_MinEccen = 21.018 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 33.142 kN.m
My = max(My,My_MinEccen) + MuaddY = 21.018 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 33.14/1050.90 = 32 mm
Actual eccenY = My / P = 0.58/1050.90 = 1 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(32,21) = 32 mm
eccenY = max(Actual eccenY,eccenYMin) = max(1,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(32 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00308 354.56 8.92 345.64 78.18 179 13.99
2 226 0.00225 336.96 8.92 328.04 74.20 90 6.64
3 226 0.00143 285.19 8.19 277.00 62.66 0 0.00
4 226 0.00060 120.18 4.55 115.63 26.15 -90 -2.34
5 226 -0.00022 -44.82 0.00 -44.82 -10.14 -179 1.81
Total231.05 20.11
xux = 380 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 380/450 = 0.304
Puc = 0.304 x 20.00 x 300 x 450
= 820.13 kN
Pux1 = Puc + Pus(Total)
= 820.13 + (231.05)
= 1051.18 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 380/450
= 0.351
Muc = 820.13 x (0.5 x 450 - 0.351 x 450) = 54.99 kN-m
Mux1 = Muc + Mus(Total)
= 54.99 + (20.11)
= 75.10 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00289 352.91 8.92 343.99 194.52 104 20.23
2 565 0.00011 22.67 0.98 21.68 12.26 -104 -1.28
Total206.78 18.95
xuy = 263 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 263/300 = 0.315
Puc = 0.315 x 20.00 x 450 x 300
= 850.50 kN
Puy1 = Puc + Pus(Total)
= 850.50 + (206.78)
= 1057.28 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 263/300
= 0.364
Muc = 850.50 x (0.5 x 300 - 0.364 x 300) = 34.70 kN-m
Muy1 = Muc + Mus(Total)
= 34.70 + (18.95)
= 53.66 kN-m
Pu/Puz = 0.675
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.675, an = 1.792
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((33.14/75.10)1.792) + ((21.02/53.66)1.792)
= 0.417 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG34 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1557.15 kN
MomentX,(Mx) = 25.79 kN-m
MomentY,(My) = 0.66 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C420.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 32.233 kN.m
My_MinEccen = 31.143 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 32.233 kN.m
My = max(My,My_MinEccen) + MuaddY = 31.143 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 25.79/1557.15 = 17 mm
Actual eccenY = My / P = 0.66/1557.15 = 0 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(17,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(0,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 12 nos. + #16 - 4 nos. (2161 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00304 354.26 8.92 345.34 138.87 177 24.58
2 226 0.00267 349.42 8.92 340.50 77.02 125 9.63
3 226 0.00231 339.06 8.92 330.14 74.67 75 5.60
4 226 0.00195 325.85 8.91 316.94 71.69 25 1.79
5 226 0.00159 302.84 8.54 294.30 66.57 -25 -1.66
6 226 0.00123 245.97 7.60 238.37 53.92 -75 -4.04
7 226 0.00087 174.09 6.07 168.01 38.00 -125 -4.75
8 402 0.00050 99.33 3.88 95.45 38.38 -177 -6.79
Total559.13 24.35
xux = 471 mm               Puc = C1.fck.B.D
ku = 1.047
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.976
C1 = 0.446 x (1 - C3/6) = 0.373
Puc = 0.373 x 20.00 x 300 x 450
= 1008.29 kN
Pux1 = Puc + Pus(Total)
= 1008.29 + (559.13)
= 1567.42 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.431
Muc = 1008.29 x (0.5 x 450 - 0.431 x 450) = 31.49 kN-m
Mux1 = Muc + Mus(Total)
= 31.49 + (24.35)
= 55.83 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1081 0.00281 352.22 8.92 343.30 371.00 102 37.84
2 1081 0.00076 152.59 5.51 147.08 158.95 -102 -16.21
Total529.96 21.63
xuy = 328 mm               Puc = C1.fck.B.D
ku = 1.094
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.843
C1 = 0.446 x (1 - C3/6) = 0.383
Puc = 0.383 x 20.00 x 450 x 300
= 1034.93 kN
Puy1 = Puc + Pus(Total)
= 1034.93 + (529.96)
= 1564.88 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.442
Muc = 1034.93 x (0.5 x 300 - 0.442 x 300) = 18.14 kN-m
Muy1 = Muc + Mus(Total)
= 18.14 + (21.63)
= 39.77 kN-m
Pu/Puz = 0.833
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.833, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((32.23/55.83)2.000) + ((31.14/39.77)2.000)
= 0.947 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG34 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 12 nos. + #16 - 4 nos. (2161 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 2003.66 kN
MomentX,(Mx) = 17.86 kN-m
MomentY,(My) = 2.68 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C420.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 41.476 kN.m
My_MinEccen = 40.073 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 41.476 kN.m
My = max(My,My_MinEccen) + MuaddY = 40.073 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 17.86/2003.66 = 9 mm
Actual eccenY = My / P = 2.68/2003.66 = 1 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(9,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(1,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 14 nos. (4398 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00309 354.70 8.92 345.78 217.26 175 38.02
2 628 0.00262 348.10 8.92 339.18 213.11 117 24.86
3 628 0.00214 332.88 8.92 323.96 203.55 58 11.87
4 628 0.00166 308.82 8.67 300.15 188.59 0 0.00
5 628 0.00119 237.58 7.45 230.13 144.60 -58 -8.43
6 628 0.00071 142.38 5.22 137.16 86.18 -117 -10.05
7 628 0.00024 47.18 1.98 45.20 28.40 -175 -4.97
Total1081.69 51.30
xux = 429 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 429/450 = 0.343
Puc = 0.343 x 20.00 x 300 x 450
= 926.44 kN
Pux1 = Puc + Pus(Total)
= 926.44 + (1081.69)
= 2008.13 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 429/450
= 0.397
Muc = 926.44 x (0.5 x 450 - 0.397 x 450) = 43.15 kN-m
Mux1 = Muc + Mus(Total)
= 43.15 + (51.30)
= 94.45 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 2199 0.00289 352.96 8.92 344.04 756.58 100 75.66
2 2199 0.00062 124.21 4.68 119.53 262.86 -100 -26.29
Total1019.43 49.37
xuy = 305 mm               Puc = C1.fck.B.D
ku = 1.016
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.083
C1 = 0.446 x (1 - C3/6) = 0.366
Puc = 0.366 x 20.00 x 450 x 300
= 986.88 kN
Puy1 = Puc + Pus(Total)
= 986.88 + (1019.43)
= 2006.31 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.421
Muc = 986.88 x (0.5 x 300 - 0.421 x 300) = 23.28 kN-m
Muy1 = Muc + Mus(Total)
= 23.28 + (49.37)
= 72.66 kN-m
Pu/Puz = 0.787
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.787, an = 1.979
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((41.48/94.45)1.979) + ((40.07/72.66)1.979)
= 0.504 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 20 = 320 mm
               iii) 300 mm
               Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG34 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 14 nos. (4398 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 2321.06 kN
MomentX,(Mx) = 7.19 kN-m
MomentY,(My) = 2.58 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C420.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 50.483 kN.m
My_MinEccen = 46.421 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 50.483 kN.m
My = max(My,My_MinEccen) + MuaddY = 46.421 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 7.19/2321.06 = 3 mm
Actual eccenY = My / P = 2.58/2321.06 = 1 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(3,22) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(1,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 16 nos. (5027 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00305 354.37 8.92 345.45 217.05 175 37.98
2 628 0.00268 349.87 8.92 340.95 214.22 125 26.78
3 628 0.00232 339.35 8.92 330.43 207.61 75 15.57
4 628 0.00195 325.80 8.91 316.89 199.11 25 4.98
5 628 0.00158 301.83 8.52 293.31 184.29 -25 -4.61
6 628 0.00121 241.98 7.53 234.45 147.31 -75 -11.05
7 628 0.00084 168.23 5.93 162.31 101.98 -125 -12.75
8 628 0.00047 94.49 3.72 90.77 57.03 -175 -9.98
Total1328.61 46.93
xux = 464 mm               Puc = C1.fck.B.D
ku = 1.031
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.027
C1 = 0.446 x (1 - C3/6) = 0.370
Puc = 0.370 x 20.00 x 300 x 450
= 998.00 kN
Pux1 = Puc + Pus(Total)
= 998.00 + (1328.61)
= 2326.60 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.426
Muc = 998.00 x (0.5 x 450 - 0.426 x 450) = 33.14 kN-m
Mux1 = Muc + Mus(Total)
= 33.14 + (46.93)
= 80.07 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 2513 0.00274 351.17 8.92 342.25 860.18 100 86.02
2 2513 0.00086 172.62 6.04 166.59 418.67 -100 -41.87
Total1278.85 44.15
xuy = 342 mm               Puc = C1.fck.B.D
ku = 1.141
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.736
C1 = 0.446 x (1 - C3/6) = 0.391
Puc = 0.391 x 20.00 x 450 x 300
= 1056.48 kN
Puy1 = Puc + Pus(Total)
= 1056.48 + (1278.85)
= 2335.33 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.450
Muc = 1056.48 x (0.5 x 300 - 0.450 x 300) = 15.83 kN-m
Muy1 = Muc + Mus(Total)
= 15.83 + (44.15)
= 59.98 kN-m
Pu/Puz = 0.849
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.849, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((50.48/80.07)2.000) + ((46.42/59.98)2.000)
= 0.997 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 20 = 320 mm
               iii) 300 mm
               Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG34 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 16 nos. (5027 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG35 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 570.79 kN
MomentX,(Mx) = 13.46 kN-m
MomentY,(My) = 43.10 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C430.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 11.815 kN.m
My_MinEccen = 11.416 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 13.459 kN.m
My = max(My,My_MinEccen) + MuaddY = 43.104 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 13.46/570.79 = 24 mm
Actual eccenY = My / P = 43.10/570.79 = 76 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(24,21) = 24 mm
eccenY = max(Actual eccenY,eccenYMin) = max(76,20) = 76 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(24 mm) > 0.05 x 300(15 mm)
and eccenY(76 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00285 352.59 8.92 343.67 77.74 179 13.91
2 226 0.00159 302.58 8.54 294.05 66.51 90 5.95
3 226 0.00032 64.54 2.65 61.89 14.00 0 0.00
4 226 -0.00094 -188.23 0.00 -188.23 -42.58 -90 3.81
5 226 -0.00221 -335.27 0.00 -335.27 -75.84 -179 13.57
Total39.83 37.25
xux = 248 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 248/450 = 0.198
Puc = 0.198 x 20.00 x 300 x 450
= 535.36 kN
Pux1 = Puc + Pus(Total)
= 535.36 + (39.83)
= 575.19 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 248/450
= 0.229
Muc = 535.36 x (0.5 x 450 - 0.229 x 450) = 65.26 kN-m
Mux1 = Muc + Mus(Total)
= 65.26 + (37.25)
= 102.51 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00257 346.80 8.92 337.88 191.06 104 19.87
2 565 -0.00166 -308.61 0.00 -308.61 -174.52 -104 18.15
Total16.55 38.02
xuy = 172 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 172/300 = 0.207
Puc = 0.207 x 20.00 x 450 x 300
= 558.14 kN
Puy1 = Puc + Pus(Total)
= 558.14 + (16.55)
= 574.69 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 172/300
= 0.239
Muc = 558.14 x (0.5 x 300 - 0.239 x 300) = 43.72 kN-m
Muy1 = Muc + Mus(Total)
= 43.72 + (38.02)
= 81.74 kN-m
Pu/Puz = 0.367
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.367, an = 1.278
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((13.46/102.51)1.278) + ((43.10/81.74)1.278)
= 0.516 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG35 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1125.22 kN
MomentX,(Mx) = 9.23 kN-m
MomentY,(My) = 40.23 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C430.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 23.292 kN.m
My_MinEccen = 22.504 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 23.292 kN.m
My = max(My,My_MinEccen) + MuaddY = 40.232 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 9.23/1125.22 = 8 mm
Actual eccenY = My / P = 40.23/1125.22 = 36 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(8,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(36,20) = 36 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(36 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00310 354.79 8.92 345.87 78.23 179 14.00
2 226 0.00233 339.75 8.92 330.83 74.83 90 6.70
3 226 0.00155 299.33 8.47 290.85 65.79 0 0.00
4 226 0.00078 155.47 5.59 149.89 33.90 -90 -3.03
5 226 0.00000 0.51 0.02 0.49 0.11 -179 -0.02
Total252.87 17.65
xux = 404 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 404/450 = 0.323
Puc = 0.323 x 20.00 x 300 x 450
= 873.28 kN
Pux1 = Puc + Pus(Total)
= 873.28 + (252.87)
= 1126.15 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 404/450
= 0.374
Muc = 873.28 x (0.5 x 450 - 0.374 x 450) = 49.61 kN-m
Mux1 = Muc + Mus(Total)
= 49.61 + (17.65)
= 67.26 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00292 353.22 8.92 344.30 194.70 104 20.25
2 565 0.00031 62.51 2.57 59.94 33.90 -104 -3.53
Total228.59 16.72
xuy = 279 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 279/300 = 0.335
Puc = 0.335 x 20.00 x 450 x 300
= 903.66 kN
Puy1 = Puc + Pus(Total)
= 903.66 + (228.59)
= 1132.25 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 279/300
= 0.387
Muc = 903.66 x (0.5 x 300 - 0.387 x 300) = 30.70 kN-m
Muy1 = Muc + Mus(Total)
= 30.70 + (16.72)
= 47.43 kN-m
Pu/Puz = 0.723
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.723, an = 1.871
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((23.29/67.26)1.871) + ((40.23/47.43)1.871)
= 0.873 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG35 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1699.03 kN
MomentX,(Mx) = 5.13 kN-m
MomentY,(My) = 38.98 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C430.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 35.170 kN.m
My_MinEccen = 33.981 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 35.170 kN.m
My = max(My,My_MinEccen) + MuaddY = 38.983 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 5.13/1699.03 = 3 mm
Actual eccenY = My / P = 38.98/1699.03 = 23 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(3,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(23,20) = 23 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(23 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 14 nos. (2815 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00311 354.88 8.92 345.96 139.12 177 24.62
2 402 0.00266 349.20 8.92 340.28 136.83 118 16.15
3 402 0.00221 335.31 8.92 326.39 131.25 59 7.74
4 402 0.00175 314.40 8.78 305.62 122.90 0 0.00
5 402 0.00130 260.13 7.83 252.31 101.46 -59 -5.99
6 402 0.00085 169.59 5.96 163.63 65.80 -118 -7.76
7 402 0.00040 79.05 3.18 75.88 30.51 -177 -5.40
Total727.87 29.36
xux = 454 mm               Puc = C1.fck.B.D
ku = 1.008
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.112
C1 = 0.446 x (1 - C3/6) = 0.363
Puc = 0.363 x 20.00 x 300 x 450
= 980.97 kN
Pux1 = Puc + Pus(Total)
= 980.97 + (727.87)
= 1708.84 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.419
Muc = 980.97 x (0.5 x 450 - 0.419 x 450) = 35.88 kN-m
Mux1 = Muc + Mus(Total)
= 35.88 + (29.36)
= 65.24 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1407 0.00283 352.39 8.92 343.47 483.41 102 49.31
2 1407 0.00073 146.64 5.34 141.30 198.87 -102 -20.28
Total682.27 29.02
xuy = 323 mm               Puc = C1.fck.B.D
ku = 1.078
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.884
C1 = 0.446 x (1 - C3/6) = 0.380
Puc = 0.380 x 20.00 x 450 x 300
= 1026.69 kN
Puy1 = Puc + Pus(Total)
= 1026.69 + (682.27)
= 1708.96 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.438
Muc = 1026.69 x (0.5 x 300 - 0.438 x 300) = 19.02 kN-m
Muy1 = Muc + Mus(Total)
= 19.02 + (29.02)
= 48.04 kN-m
Pu/Puz = 0.822
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.822, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((35.17/65.24)2.000) + ((38.98/48.04)2.000)
= 0.949 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG35 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 14 nos. (2815 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 2307.45 kN
MomentX,(Mx) = 2.27 kN-m
MomentY,(My) = 36.62 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C430.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 47.764 kN.m
My_MinEccen = 46.149 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 47.764 kN.m
My = max(My,My_MinEccen) + MuaddY = 46.149 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 2.27/2307.45 = 1 mm
Actual eccenY = My / P = 36.62/2307.45 = 16 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(1,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(16,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 16 nos. (5027 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00305 354.37 8.92 345.45 217.05 175 37.98
2 628 0.00268 349.87 8.92 340.95 214.22 125 26.78
3 628 0.00232 339.35 8.92 330.43 207.61 75 15.57
4 628 0.00195 325.80 8.91 316.89 199.11 25 4.98
5 628 0.00158 301.83 8.52 293.31 184.29 -25 -4.61
6 628 0.00121 241.98 7.53 234.45 147.31 -75 -11.05
7 628 0.00084 168.23 5.93 162.31 101.98 -125 -12.75
8 628 0.00047 94.49 3.72 90.77 57.03 -175 -9.98
Total1328.61 46.93
xux = 464 mm               Puc = C1.fck.B.D
ku = 1.031
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.027
C1 = 0.446 x (1 - C3/6) = 0.370
Puc = 0.370 x 20.00 x 300 x 450
= 998.00 kN
Pux1 = Puc + Pus(Total)
= 998.00 + (1328.61)
= 2326.60 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.426
Muc = 998.00 x (0.5 x 450 - 0.426 x 450) = 33.14 kN-m
Mux1 = Muc + Mus(Total)
= 33.14 + (46.93)
= 80.07 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 2513 0.00275 351.60 8.92 342.68 861.24 100 86.12
2 2513 0.00084 167.52 5.91 161.61 406.18 -100 -40.62
Total1267.42 45.51
xuy = 338 mm               Puc = C1.fck.B.D
ku = 1.125
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.769
C1 = 0.446 x (1 - C3/6) = 0.389
Puc = 0.389 x 20.00 x 450 x 300
= 1049.78 kN
Puy1 = Puc + Pus(Total)
= 1049.78 + (1267.42)
= 2317.20 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.447
Muc = 1049.78 x (0.5 x 300 - 0.447 x 300) = 16.55 kN-m
Muy1 = Muc + Mus(Total)
= 16.55 + (45.51)
= 62.05 kN-m
Pu/Puz = 0.844
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.844, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((47.76/80.07)2.000) + ((46.15/62.05)2.000)
= 0.909 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 20 = 320 mm
               iii) 300 mm
               Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG35 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 16 nos. (5027 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 3000.73 kN
MomentX,(Mx) = 0.09 kN-m
MomentY,(My) = 22.59 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C430.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 65.266 kN.m
My_MinEccen = 60.015 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 65.266 kN.m
My = max(My,My_MinEccen) + MuaddY = 60.015 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 0.09/3000.73 = 0 mm
Actual eccenY = My / P = 22.59/3000.73 = 8 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(0,22) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(8,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #32 - 10 nos. (8042 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1608 0.00306 354.46 8.92 345.54 555.81 169 93.93
2 1608 0.00241 342.70 8.92 333.78 536.88 85 45.37
3 1608 0.00175 314.19 8.78 305.41 491.25 0 0.00
4 1608 0.00109 218.56 7.08 211.47 340.15 -85 -28.74
5 1608 0.00044 87.11 3.46 83.65 134.55 -169 -22.74
Total2058.63 87.82
xux = 450 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 450/450 = 0.360
Puc = 0.360 x 20.00 x 300 x 450
= 972.00 kN
Pux1 = Puc + Pus(Total)
= 972.00 + (2058.63)
= 3030.63 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 450/450
= 0.416
Muc = 972.00 x (0.5 x 450 - 0.416 x 450) = 36.74 kN-m
Mux1 = Muc + Mus(Total)
= 36.74 + (87.82)
= 124.56 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 4021 0.00276 351.83 8.92 342.91 1378.91 94 129.62
2 4021 0.00079 157.22 5.63 151.59 609.57 -94 -57.30
Total1988.48 72.32
xuy = 319 mm               Puc = C1.fck.B.D
ku = 1.063
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.929
C1 = 0.446 x (1 - C3/6) = 0.377
Puc = 0.377 x 20.00 x 450 x 300
= 1017.83 kN
Puy1 = Puc + Pus(Total)
= 1017.83 + (1988.48)
= 3006.31 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.435
Muc = 1017.83 x (0.5 x 300 - 0.435 x 300) = 19.97 kN-m
Muy1 = Muc + Mus(Total)
= 19.97 + (72.32)
= 92.29 kN-m
Pu/Puz = 0.823
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.823, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((65.27/124.56)2.000) + ((60.01/92.29)2.000)
= 0.697 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 32/4 = 8 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (8, 6) = 8 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 32 = 512 mm
               iii) 300 mm
               Required spacing = minimum of (300, 512, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG35 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #32 - 10 nos. (8042 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG36 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 420.60 kN
MomentX,(Mx) = 6.08 kN-m
MomentY,(My) = 86.78 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C440.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 8.706 kN.m
My_MinEccen = 8.412 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 8.706 kN.m
My = max(My,My_MinEccen) + MuaddY = 86.779 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 6.08/420.60 = 14 mm
Actual eccenY = My / P = 86.78/420.60 = 206 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(14,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(206,20) = 206 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(206 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 6 nos. + #16 - 4 nos. (1483 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00270 350.35 8.92 341.43 137.30 177 24.30
2 226 0.00122 243.70 7.56 236.15 53.41 88 4.67
3 226 -0.00023 -46.67 0.00 -46.67 -10.56 0 0.00
4 226 -0.00169 -310.14 0.00 -310.14 -70.15 -88 6.14
5 402 -0.00317 -355.39 0.00 -355.39 -142.91 -177 25.30
Total-32.91 60.41
xux = 211 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 211/450 = 0.169
Puc = 0.169 x 20.00 x 300 x 450
= 455.63 kN
Pux1 = Puc + Pus(Total)
= 455.63 + (-32.91)
= 422.72 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 211/450
= 0.195
Muc = 455.63 x (0.5 x 450 - 0.195 x 450) = 62.53 kN-m
Mux1 = Muc + Mus(Total)
= 62.53 + (60.41)
= 122.94 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 741 0.00226 337.44 8.92 328.52 243.57 102 24.84
2 741 -0.00299 -353.80 0.00 -353.80 -262.31 -102 26.76
Total-18.74 51.60
xuy = 136 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 136/300 = 0.163
Puc = 0.163 x 20.00 x 450 x 300
= 440.44 kN
Puy1 = Puc + Pus(Total)
= 440.44 + (-18.74)
= 421.70 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 136/300
= 0.189
Muc = 440.44 x (0.5 x 300 - 0.189 x 300) = 41.16 kN-m
Muy1 = Muc + Mus(Total)
= 41.16 + (51.60)
= 92.76 kN-m
Pu/Puz = 0.253
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.253, an = 1.088
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((8.71/122.94)1.088) + ((86.78/92.76)1.088)
= 0.986 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG36 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 6 nos. + #16 - 4 nos. (1483 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 859.35 kN
MomentX,(Mx) = 4.53 kN-m
MomentY,(My) = 79.73 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C440.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 17.789 kN.m
My_MinEccen = 17.187 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 17.789 kN.m
My = max(My,My_MinEccen) + MuaddY = 79.732 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 4.53/859.35 = 5 mm
Actual eccenY = My / P = 79.73/859.35 = 93 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(5,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(93,20) = 93 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(93 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 16 nos. (1810 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00296 353.59 8.92 344.67 77.96 179 13.96
2 226 0.00237 341.29 8.92 332.37 75.18 128 9.61
3 226 0.00177 315.65 8.81 306.84 69.41 77 5.32
4 226 0.00118 235.57 7.41 228.16 51.61 26 1.32
5 226 0.00058 116.47 4.44 112.03 25.34 -26 -0.65
6 226 -0.00001 -2.63 0.00 -2.63 -0.59 -77 0.05
7 226 -0.00061 -121.73 0.00 -121.73 -27.53 -128 3.52
8 226 -0.00120 -240.83 0.00 -240.83 -54.47 -179 9.75
Total216.90 42.88
xux = 301 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 301/450 = 0.240
Puc = 0.240 x 20.00 x 300 x 450
= 649.27 kN
Pux1 = Puc + Pus(Total)
= 649.27 + (216.90)
= 866.16 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 301/450
= 0.278
Muc = 649.27 x (0.5 x 450 - 0.278 x 450) = 64.90 kN-m
Mux1 = Muc + Mus(Total)
= 64.90 + (42.88)
= 107.78 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 905 0.00274 351.20 8.92 342.28 309.69 104 32.21
2 905 -0.00071 -142.90 0.00 -142.90 -129.30 -104 13.45
Total180.39 45.65
xuy = 211 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 211/300 = 0.253
Puc = 0.253 x 20.00 x 450 x 300
= 683.44 kN
Puy1 = Puc + Pus(Total)
= 683.44 + (180.39)
= 863.83 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 211/300
= 0.293
Muc = 683.44 x (0.5 x 300 - 0.293 x 300) = 42.54 kN-m
Muy1 = Muc + Mus(Total)
= 42.54 + (45.65)
= 88.20 kN-m
Pu/Puz = 0.488
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.488, an = 1.480
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((17.79/107.78)1.480) + ((79.73/88.20)1.480)
= 0.931 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG36 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 16 nos. (1810 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1295.73 kN
MomentX,(Mx) = 3.79 kN-m
MomentY,(My) = 77.62 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C440.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 26.822 kN.m
My_MinEccen = 25.915 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 26.822 kN.m
My = max(My,My_MinEccen) + MuaddY = 77.622 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 3.79/1295.73 = 3 mm
Actual eccenY = My / P = 77.62/1295.73 = 60 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(3,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(60,20) = 60 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(60 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 14 nos. (2815 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00303 354.16 8.92 345.24 138.83 177 24.57
2 402 0.00245 343.84 8.92 334.92 134.68 118 15.89
3 402 0.00187 321.79 8.88 312.91 125.83 59 7.42
4 402 0.00129 258.62 7.81 250.82 100.86 0 0.00
5 402 0.00071 142.88 5.23 137.65 55.35 -59 -3.27
6 402 0.00014 27.14 1.17 25.97 10.44 -118 -1.23
7 402 -0.00044 -88.60 0.00 -88.60 -35.63 -177 6.31
Total530.36 49.70
xux = 357 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 357/450 = 0.285
Puc = 0.285 x 20.00 x 300 x 450
= 770.77 kN
Pux1 = Puc + Pus(Total)
= 770.77 + (530.36)
= 1301.13 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 357/450
= 0.330
Muc = 770.77 x (0.5 x 450 - 0.330 x 450) = 59.01 kN-m
Mux1 = Muc + Mus(Total)
= 59.01 + (49.70)
= 108.70 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1407 0.00284 352.47 8.92 343.55 483.52 102 49.32
2 1407 0.00002 3.11 0.14 2.97 4.18 -102 -0.43
Total487.70 48.89
xuy = 253 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 253/300 = 0.304
Puc = 0.304 x 20.00 x 450 x 300
= 820.13 kN
Puy1 = Puc + Pus(Total)
= 820.13 + (487.70)
= 1307.83 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 253/300
= 0.351
Muc = 820.13 x (0.5 x 300 - 0.351 x 300) = 36.66 kN-m
Muy1 = Muc + Mus(Total)
= 36.66 + (48.89)
= 85.55 kN-m
Pu/Puz = 0.627
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.627, an = 1.712
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((26.82/108.70)1.712) + ((77.62/85.55)1.712)
= 0.938 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG36 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 14 nos. (2815 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1728.61 kN
MomentX,(Mx) = 3.03 kN-m
MomentY,(My) = 74.41 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C440.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 35.782 kN.m
My_MinEccen = 34.572 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 35.782 kN.m
My = max(My,My_MinEccen) + MuaddY = 74.407 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 3.03/1728.61 = 2 mm
Actual eccenY = My / P = 74.41/1728.61 = 43 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(2,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(43,20) = 43 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(43 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 14 nos. (4398 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00304 354.22 8.92 345.30 216.96 175 37.97
2 628 0.00250 345.03 8.92 336.11 211.18 117 24.64
3 628 0.00196 326.14 8.92 317.23 199.32 58 11.63
4 628 0.00142 283.26 8.16 275.10 172.85 0 0.00
5 628 0.00088 175.21 6.10 169.11 106.25 -58 -6.20
6 628 0.00034 67.17 2.74 64.42 40.48 -117 -4.72
7 628 -0.00020 -40.88 0.00 -40.88 -25.68 -175 4.49
Total921.36 67.81
xux = 378 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 378/450 = 0.302
Puc = 0.302 x 20.00 x 300 x 450
= 816.33 kN
Pux1 = Puc + Pus(Total)
= 816.33 + (921.36)
= 1737.69 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 378/450
= 0.349
Muc = 816.33 x (0.5 x 450 - 0.349 x 450) = 55.33 kN-m
Mux1 = Muc + Mus(Total)
= 55.33 + (67.81)
= 123.14 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 2199 0.00285 352.59 8.92 343.67 755.78 100 75.58
2 2199 0.00025 50.72 2.12 48.61 106.89 -100 -10.69
Total862.67 64.89
xuy = 270 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 270/300 = 0.323
Puc = 0.323 x 20.00 x 450 x 300
= 873.28 kN
Puy1 = Puc + Pus(Total)
= 873.28 + (862.67)
= 1735.95 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 270/300
= 0.374
Muc = 873.28 x (0.5 x 300 - 0.374 x 300) = 33.08 kN-m
Muy1 = Muc + Mus(Total)
= 33.08 + (64.89)
= 97.96 kN-m
Pu/Puz = 0.679
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.679, an = 1.799
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((35.78/123.14)1.799) + ((74.41/97.96)1.799)
= 0.718 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 20 = 320 mm
               iii) 300 mm
               Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG36 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 14 nos. (4398 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 2152.66 kN
MomentX,(Mx) = 0.67 kN-m
MomentY,(My) = 49.23 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C440.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 46.820 kN.m
My_MinEccen = 43.053 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 46.820 kN.m
My = max(My,My_MinEccen) + MuaddY = 49.227 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 0.67/2152.66 = 0 mm
Actual eccenY = My / P = 49.23/2152.66 = 23 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(0,22) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(23,20) = 23 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(23 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 16 nos. (5027 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00309 354.70 8.92 345.78 217.26 175 38.02
2 628 0.00268 349.84 8.92 340.92 214.21 125 26.78
3 628 0.00228 337.88 8.92 328.96 206.69 75 15.50
4 628 0.00187 321.55 8.88 312.67 196.46 25 4.91
5 628 0.00146 290.59 8.27 282.32 177.39 -25 -4.43
6 628 0.00105 210.38 6.92 203.47 127.84 -75 -9.59
7 628 0.00064 128.78 4.82 123.96 77.89 -125 -9.74
8 628 0.00024 47.18 1.98 45.20 28.40 -175 -4.97
Total1246.13 56.48
xux = 429 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 429/450 = 0.343
Puc = 0.343 x 20.00 x 300 x 450
= 926.44 kN
Pux1 = Puc + Pus(Total)
= 926.44 + (1246.13)
= 2172.57 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 429/450
= 0.397
Muc = 926.44 x (0.5 x 450 - 0.397 x 450) = 43.15 kN-m
Mux1 = Muc + Mus(Total)
= 43.15 + (56.48)
= 99.63 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 2513 0.00288 352.85 8.92 343.93 864.40 100 86.44
2 2513 0.00064 127.83 4.79 123.04 309.23 -100 -30.92
Total1173.63 55.52
xuy = 307 mm               Puc = C1.fck.B.D
ku = 1.023
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.055
C1 = 0.446 x (1 - C3/6) = 0.368
Puc = 0.368 x 20.00 x 450 x 300
= 992.55 kN
Puy1 = Puc + Pus(Total)
= 992.55 + (1173.63)
= 2166.18 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.424
Muc = 992.55 x (0.5 x 300 - 0.424 x 300) = 22.68 kN-m
Muy1 = Muc + Mus(Total)
= 22.68 + (55.52)
= 78.19 kN-m
Pu/Puz = 0.787
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.787, an = 1.979
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((46.82/99.63)1.979) + ((49.23/78.19)1.979)
= 0.625 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 20 = 320 mm
               iii) 300 mm
               Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG36 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 16 nos. (5027 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG37 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 435.87 kN
MomentX,(Mx) = 12.26 kN-m
MomentY,(My) = 94.68 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C450.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 9.022 kN.m
My_MinEccen = 8.717 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 12.263 kN.m
My = max(My,My_MinEccen) + MuaddY = 94.679 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 12.26/435.87 = 28 mm
Actual eccenY = My / P = 94.68/435.87 = 217 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(28,21) = 28 mm
eccenY = max(Actual eccenY,eccenYMin) = max(217,20) = 217 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(28 mm) > 0.05 x 300(15 mm)
and eccenY(217 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 16 nos. (1810 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00276 351.76 8.92 342.84 77.55 179 13.88
2 226 0.00193 325.31 8.91 316.40 71.57 128 9.15
3 226 0.00111 221.86 7.15 214.71 48.57 77 3.73
4 226 0.00028 56.95 2.36 54.59 12.35 26 0.32
5 226 -0.00054 -107.96 0.00 -107.96 -24.42 -26 0.62
6 226 -0.00136 -272.87 0.00 -272.87 -61.72 -77 4.73
7 226 -0.00219 -334.68 0.00 -334.68 -75.70 -128 9.68
8 226 -0.00301 -354.02 0.00 -354.02 -80.08 -179 14.33
Total-31.89 56.45
xux = 217 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 217/450 = 0.174
Puc = 0.174 x 20.00 x 300 x 450
= 468.91 kN
Pux1 = Puc + Pus(Total)
= 468.91 + (-31.89)
= 437.02 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 217/450
= 0.201
Muc = 468.91 x (0.5 x 450 - 0.201 x 450) = 63.16 kN-m
Mux1 = Muc + Mus(Total)
= 63.16 + (56.45)
= 119.60 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 905 0.00236 340.78 8.92 331.86 300.26 104 31.23
2 905 -0.00282 -352.34 0.00 -352.34 -318.79 -104 33.15
Total-18.53 64.38
xuy = 141 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 141/300 = 0.169
Puc = 0.169 x 20.00 x 450 x 300
= 455.63 kN
Puy1 = Puc + Pus(Total)
= 455.63 + (-18.53)
= 437.10 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 141/300
= 0.195
Muc = 455.63 x (0.5 x 300 - 0.195 x 300) = 41.69 kN-m
Muy1 = Muc + Mus(Total)
= 41.69 + (64.38)
= 106.07 kN-m
Pu/Puz = 0.247
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.247, an = 1.079
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((12.26/119.60)1.079) + ((94.68/106.07)1.079)
= 0.970 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG37 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 16 nos. (1810 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 889.46 kN
MomentX,(Mx) = 11.63 kN-m
MomentY,(My) = 86.07 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C450.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 18.412 kN.m
My_MinEccen = 17.789 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 18.412 kN.m
My = max(My,My_MinEccen) + MuaddY = 86.067 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 11.63/889.46 = 13 mm
Actual eccenY = My / P = 86.07/889.46 = 97 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(13,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(97,20) = 97 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(97 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 4 nos. + #16 - 8 nos. (2061 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00295 353.44 8.92 344.52 138.54 177 24.52
2 402 0.00211 331.94 8.92 323.02 129.89 105 13.59
3 226 0.00130 260.80 7.84 252.96 57.22 34 1.96
4 226 0.00052 103.36 4.01 99.34 22.47 -34 -0.77
5 402 -0.00029 -58.69 0.00 -58.69 -23.60 -105 2.47
6 402 -0.00113 -225.35 0.00 -225.35 -90.62 -177 16.04
Total233.90 57.80
xux = 304 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 304/450 = 0.243
Puc = 0.243 x 20.00 x 300 x 450
= 656.86 kN
Pux1 = Puc + Pus(Total)
= 656.86 + (233.90)
= 890.76 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 304/450
= 0.281
Muc = 656.86 x (0.5 x 450 - 0.281 x 450) = 64.70 kN-m
Mux1 = Muc + Mus(Total)
= 64.70 + (57.80)
= 122.50 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1030 0.00270 350.35 8.92 341.43 351.82 102 35.89
2 1030 -0.00068 -136.27 0.00 -136.27 -140.41 -102 14.32
Total211.41 50.21
xuy = 211 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 211/300 = 0.253
Puc = 0.253 x 20.00 x 450 x 300
= 683.44 kN
Puy1 = Puc + Pus(Total)
= 683.44 + (211.41)
= 894.85 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 211/300
= 0.293
Muc = 683.44 x (0.5 x 300 - 0.293 x 300) = 42.54 kN-m
Muy1 = Muc + Mus(Total)
= 42.54 + (50.21)
= 92.75 kN-m
Pu/Puz = 0.484
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.484, an = 1.473
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((18.41/122.50)1.473) + ((86.07/92.75)1.473)
= 0.957 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG37 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 4 nos. + #16 - 8 nos. (2061 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1340.18 kN
MomentX,(Mx) = 9.61 kN-m
MomentY,(My) = 83.25 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C450.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 27.742 kN.m
My_MinEccen = 26.804 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 27.742 kN.m
My = max(My,My_MinEccen) + MuaddY = 83.253 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 9.61/1340.18 = 7 mm
Actual eccenY = My / P = 83.25/1340.18 = 62 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(7,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(62,20) = 62 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(62 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 16 nos. (3217 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00302 354.07 8.92 345.15 138.79 177 24.57
2 402 0.00251 345.47 8.92 336.55 135.33 126 17.11
3 402 0.00201 328.02 8.92 319.10 128.32 76 9.73
4 402 0.00150 294.55 8.37 286.18 115.08 25 2.91
5 402 0.00100 199.15 6.67 192.48 77.40 -25 -1.96
6 402 0.00049 97.95 3.83 94.12 37.85 -76 -2.87
7 402 -0.00002 -3.25 0.00 -3.25 -1.31 -126 0.17
8 402 -0.00052 -104.45 0.00 -104.45 -42.00 -177 7.43
Total589.47 57.09
xux = 350 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 350/450 = 0.280
Puc = 0.280 x 20.00 x 300 x 450
= 755.58 kN
Pux1 = Puc + Pus(Total)
= 755.58 + (589.47)
= 1345.04 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 350/450
= 0.323
Muc = 755.58 x (0.5 x 450 - 0.323 x 450) = 60.05 kN-m
Mux1 = Muc + Mus(Total)
= 60.05 + (57.09)
= 117.15 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1608 0.00282 352.36 8.92 343.44 552.42 102 56.35
2 1608 -0.00005 -10.04 0.00 -10.04 -16.15 -102 1.65
Total536.27 57.99
xuy = 248 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 248/300 = 0.298
Puc = 0.298 x 20.00 x 450 x 300
= 804.94 kN
Puy1 = Puc + Pus(Total)
= 804.94 + (536.27)
= 1341.21 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 248/300
= 0.345
Muc = 804.94 x (0.5 x 300 - 0.345 x 300) = 37.55 kN-m
Muy1 = Muc + Mus(Total)
= 37.55 + (57.99)
= 95.54 kN-m
Pu/Puz = 0.613
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.613, an = 1.688
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((27.74/117.15)1.688) + ((83.25/95.54)1.688)
= 0.881 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG37 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 16 nos. (3217 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1786.69 kN
MomentX,(Mx) = 6.69 kN-m
MomentY,(My) = 79.11 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C450.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 36.985 kN.m
My_MinEccen = 35.734 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 36.985 kN.m
My = max(My,My_MinEccen) + MuaddY = 79.107 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 6.69/1786.69 = 4 mm
Actual eccenY = My / P = 79.11/1786.69 = 44 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(4,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(44,20) = 44 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(44 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 14 nos. (4398 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00305 354.32 8.92 345.40 217.02 175 37.98
2 628 0.00252 345.62 8.92 336.70 211.55 117 24.68
3 628 0.00199 327.43 8.92 318.51 200.13 58 11.67
4 628 0.00146 290.94 8.28 282.66 177.60 0 0.00
5 628 0.00094 187.14 6.39 180.74 113.56 -58 -6.62
6 628 0.00041 81.55 3.27 78.28 49.19 -117 -5.74
7 628 -0.00012 -24.04 0.00 -24.04 -15.11 -175 2.64
Total953.94 64.61
xux = 387 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 387/450 = 0.309
Puc = 0.309 x 20.00 x 300 x 450
= 835.31 kN
Pux1 = Puc + Pus(Total)
= 835.31 + (953.94)
= 1789.26 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 387/450
= 0.358
Muc = 835.31 x (0.5 x 450 - 0.358 x 450) = 53.56 kN-m
Mux1 = Muc + Mus(Total)
= 53.56 + (64.61)
= 118.18 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 2199 0.00287 352.74 8.92 343.82 756.10 100 75.61
2 2199 0.00034 67.23 2.75 64.49 141.81 -100 -14.18
Total897.91 61.43
xuy = 277 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 277/300 = 0.332
Puc = 0.332 x 20.00 x 450 x 300
= 896.06 kN
Puy1 = Puc + Pus(Total)
= 896.06 + (897.91)
= 1793.97 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 277/300
= 0.384
Muc = 896.06 x (0.5 x 300 - 0.384 x 300) = 31.32 kN-m
Muy1 = Muc + Mus(Total)
= 31.32 + (61.43)
= 92.75 kN-m
Pu/Puz = 0.702
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.702, an = 1.837
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((36.98/118.18)1.837) + ((79.11/92.75)1.837)
= 0.865 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 20 = 320 mm
               iii) 300 mm
               Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG37 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 14 nos. (4398 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 2222.12 kN
MomentX,(Mx) = 3.04 kN-m
MomentY,(My) = 51.49 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C450.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 48.331 kN.m
My_MinEccen = 44.442 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 48.331 kN.m
My = max(My,My_MinEccen) + MuaddY = 51.493 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 3.04/2222.12 = 1 mm
Actual eccenY = My / P = 51.49/2222.12 = 23 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(1,22) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(23,20) = 23 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(23 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 16 nos. (5027 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00310 354.82 8.92 345.90 217.33 175 38.03
2 628 0.00271 350.51 8.92 341.59 214.63 125 26.83
3 628 0.00231 339.30 8.92 330.38 207.59 75 15.57
4 628 0.00192 324.78 8.91 315.88 198.47 25 4.96
5 628 0.00152 296.72 8.42 288.31 181.15 -25 -4.53
6 628 0.00113 225.93 7.23 218.70 137.41 -75 -10.31
7 628 0.00073 146.91 5.35 141.56 88.95 -125 -11.12
8 628 0.00034 67.90 2.77 65.13 40.92 -175 -7.16
Total1286.45 52.28
xux = 443 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 443/450 = 0.354
Puc = 0.354 x 20.00 x 300 x 450
= 956.81 kN
Pux1 = Puc + Pus(Total)
= 956.81 + (1286.45)
= 2243.26 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 443/450
= 0.410
Muc = 956.81 x (0.5 x 450 - 0.410 x 450) = 38.97 kN-m
Mux1 = Muc + Mus(Total)
= 38.97 + (52.28)
= 91.24 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 2513 0.00283 352.38 8.92 343.46 863.21 100 86.32
2 2513 0.00072 144.60 5.28 139.32 350.14 -100 -35.01
Total1213.35 51.31
xuy = 319 mm               Puc = C1.fck.B.D
ku = 1.063
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.929
C1 = 0.446 x (1 - C3/6) = 0.377
Puc = 0.377 x 20.00 x 450 x 300
= 1017.83 kN
Puy1 = Puc + Pus(Total)
= 1017.83 + (1213.35)
= 2231.18 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.435
Muc = 1017.83 x (0.5 x 300 - 0.435 x 300) = 19.97 kN-m
Muy1 = Muc + Mus(Total)
= 19.97 + (51.31)
= 71.28 kN-m
Pu/Puz = 0.813
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.813, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((48.33/91.24)2.000) + ((51.49/71.28)2.000)
= 0.803 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 20 = 320 mm
               iii) 300 mm
               Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG37 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 16 nos. (5027 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG38 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 565.62 kN
MomentX,(Mx) = 23.45 kN-m
MomentY,(My) = 58.39 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C460.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 11.708 kN.m
My_MinEccen = 11.312 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 23.450 kN.m
My = max(My,My_MinEccen) + MuaddY = 58.386 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 23.45/565.62 = 41 mm
Actual eccenY = My / P = 58.39/565.62 = 103 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(41,21) = 41 mm
eccenY = max(Actual eccenY,eccenYMin) = max(103,20) = 103 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(41 mm) > 0.05 x 300(15 mm)
and eccenY(103 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00285 352.55 8.92 343.63 77.73 179 13.91
2 226 0.00157 301.29 8.51 292.78 66.22 90 5.93
3 226 0.00030 60.00 2.48 57.52 13.01 0 0.00
4 226 -0.00097 -194.58 0.00 -194.58 -44.01 -90 3.94
5 226 -0.00225 -336.77 0.00 -336.77 -76.17 -179 13.64
Total36.78 37.41
xux = 246 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 246/450 = 0.197
Puc = 0.197 x 20.00 x 300 x 450
= 531.56 kN
Pux1 = Puc + Pus(Total)
= 531.56 + (36.78)
= 568.34 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 246/450
= 0.228
Muc = 531.56 x (0.5 x 450 - 0.228 x 450) = 65.18 kN-m
Mux1 = Muc + Mus(Total)
= 65.18 + (37.41)
= 102.60 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00256 346.63 8.92 337.71 190.97 104 19.86
2 565 -0.00170 -310.82 0.00 -310.82 -175.76 -104 18.28
Total15.21 38.14
xuy = 171 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 171/300 = 0.205
Puc = 0.205 x 20.00 x 450 x 300
= 554.34 kN
Puy1 = Puc + Pus(Total)
= 554.34 + (15.21)
= 569.55 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 171/300
= 0.237
Muc = 554.34 x (0.5 x 300 - 0.237 x 300) = 43.70 kN-m
Muy1 = Muc + Mus(Total)
= 43.70 + (38.14)
= 81.84 kN-m
Pu/Puz = 0.363
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.363, an = 1.272
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((23.45/102.60)1.272) + ((58.39/81.84)1.272)
= 0.804 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG38 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1110.02 kN
MomentX,(Mx) = 20.01 kN-m
MomentY,(My) = 54.57 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C460.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 22.977 kN.m
My_MinEccen = 22.200 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 22.977 kN.m
My = max(My,My_MinEccen) + MuaddY = 54.566 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 20.01/1110.02 = 18 mm
Actual eccenY = My / P = 54.57/1110.02 = 49 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(18,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(49,20) = 49 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(49 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 6 nos. + #16 - 4 nos. (1483 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00306 354.44 8.92 345.52 138.94 177 24.59
2 226 0.00224 336.71 8.92 327.79 74.14 88 6.49
3 226 0.00144 289.17 8.23 280.94 63.55 0 0.00
4 226 0.00065 129.15 4.83 124.32 28.12 -88 -2.46
5 402 -0.00017 -34.34 0.00 -34.34 -13.81 -177 2.44
Total290.95 31.06
xux = 383 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 383/450 = 0.307
Puc = 0.307 x 20.00 x 300 x 450
= 827.72 kN
Pux1 = Puc + Pus(Total)
= 827.72 + (290.95)
= 1118.66 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 383/450
= 0.354
Muc = 827.72 x (0.5 x 450 - 0.354 x 450) = 54.29 kN-m
Mux1 = Muc + Mus(Total)
= 54.29 + (31.06)
= 85.35 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 741 0.00285 352.62 8.92 343.70 254.83 102 25.99
2 741 0.00011 21.95 0.95 20.99 15.57 -102 -1.59
Total270.39 24.40
xuy = 260 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 260/300 = 0.312
Puc = 0.312 x 20.00 x 450 x 300
= 842.91 kN
Puy1 = Puc + Pus(Total)
= 842.91 + (270.39)
= 1113.30 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 260/300
= 0.361
Muc = 842.91 x (0.5 x 300 - 0.361 x 300) = 35.21 kN-m
Muy1 = Muc + Mus(Total)
= 35.21 + (24.40)
= 59.62 kN-m
Pu/Puz = 0.667
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.667, an = 1.779
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((22.98/85.35)1.779) + ((54.57/59.62)1.779)
= 0.951 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG38 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 6 nos. + #16 - 4 nos. (1483 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1681.31 kN
MomentX,(Mx) = 14.56 kN-m
MomentY,(My) = 51.65 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C460.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 34.803 kN.m
My_MinEccen = 33.626 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 34.803 kN.m
My = max(My,My_MinEccen) + MuaddY = 51.650 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 14.56/1681.31 = 9 mm
Actual eccenY = My / P = 51.65/1681.31 = 31 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(9,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(31,20) = 31 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(31 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 16 nos. (3217 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00310 354.79 8.92 345.87 139.08 177 24.62
2 402 0.00268 349.80 8.92 340.88 137.08 126 17.33
3 402 0.00226 337.39 8.92 328.47 132.08 76 10.02
4 402 0.00184 320.00 8.87 311.14 125.12 25 3.16
5 402 0.00142 284.71 8.18 276.53 111.20 -25 -2.81
6 402 0.00100 200.80 6.71 194.09 78.05 -76 -5.92
7 402 0.00058 116.89 4.45 112.44 45.21 -126 -5.72
8 402 0.00016 32.98 1.41 31.57 12.69 -177 -2.25
Total780.52 38.44
xux = 422 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 422/450 = 0.338
Puc = 0.338 x 20.00 x 300 x 450
= 911.25 kN
Pux1 = Puc + Pus(Total)
= 911.25 + (780.52)
= 1691.77 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 422/450
= 0.390
Muc = 911.25 x (0.5 x 450 - 0.390 x 450) = 45.11 kN-m
Mux1 = Muc + Mus(Total)
= 45.11 + (38.44)
= 83.54 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1608 0.00294 353.34 8.92 344.42 553.99 102 56.51
2 1608 0.00054 107.37 4.15 103.22 166.04 -102 -16.94
Total720.03 39.57
xuy = 298 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 298/300 = 0.357
Puc = 0.357 x 20.00 x 450 x 300
= 964.41 kN
Puy1 = Puc + Pus(Total)
= 964.41 + (720.03)
= 1684.43 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 298/300
= 0.413
Muc = 964.41 x (0.5 x 300 - 0.413 x 300) = 25.24 kN-m
Muy1 = Muc + Mus(Total)
= 25.24 + (39.57)
= 64.81 kN-m
Pu/Puz = 0.769
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.769, an = 1.948
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((34.80/83.54)1.948) + ((51.65/64.81)1.948)
= 0.824 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG38 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 16 nos. (3217 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 2298.28 kN
MomentX,(Mx) = 8.42 kN-m
MomentY,(My) = 46.53 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C460.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 47.574 kN.m
My_MinEccen = 45.966 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 47.574 kN.m
My = max(My,My_MinEccen) + MuaddY = 46.528 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 8.42/2298.28 = 4 mm
Actual eccenY = My / P = 46.53/2298.28 = 20 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(4,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(20,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 16 nos. (5027 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00308 354.61 8.92 345.69 217.21 175 38.01
2 628 0.00270 350.33 8.92 341.41 214.52 125 26.81
3 628 0.00232 339.66 8.92 330.74 207.81 75 15.59
4 628 0.00195 325.75 8.91 316.84 199.08 25 4.98
5 628 0.00157 300.77 8.50 292.27 183.64 -25 -4.59
6 628 0.00119 237.77 7.45 230.32 144.71 -75 -10.85
7 628 0.00081 162.06 5.76 156.30 98.20 -125 -12.28
8 628 0.00043 86.35 3.44 82.92 52.10 -175 -9.12
Total1317.26 48.55
xux = 457 mm               Puc = C1.fck.B.D
ku = 1.016
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.083
C1 = 0.446 x (1 - C3/6) = 0.366
Puc = 0.366 x 20.00 x 300 x 450
= 986.88 kN
Pux1 = Puc + Pus(Total)
= 986.88 + (1317.26)
= 2304.13 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.421
Muc = 986.88 x (0.5 x 450 - 0.421 x 450) = 34.93 kN-m
Mux1 = Muc + Mus(Total)
= 34.93 + (48.55)
= 83.48 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 2513 0.00275 351.60 8.92 342.68 861.24 100 86.12
2 2513 0.00084 167.52 5.91 161.61 406.18 -100 -40.62
Total1267.42 45.51
xuy = 338 mm               Puc = C1.fck.B.D
ku = 1.125
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.769
C1 = 0.446 x (1 - C3/6) = 0.389
Puc = 0.389 x 20.00 x 450 x 300
= 1049.78 kN
Puy1 = Puc + Pus(Total)
= 1049.78 + (1267.42)
= 2317.20 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.447
Muc = 1049.78 x (0.5 x 300 - 0.447 x 300) = 16.55 kN-m
Muy1 = Muc + Mus(Total)
= 16.55 + (45.51)
= 62.05 kN-m
Pu/Puz = 0.841
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.841, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((47.57/83.48)2.000) + ((46.53/62.05)2.000)
= 0.887 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 20 = 320 mm
               iii) 300 mm
               Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG38 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 16 nos. (5027 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 3011.29 kN
MomentX,(Mx) = 2.68 kN-m
MomentY,(My) = 28.03 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C460.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 65.495 kN.m
My_MinEccen = 60.226 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 65.495 kN.m
My = max(My,My_MinEccen) + MuaddY = 60.226 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 2.68/3011.29 = 1 mm
Actual eccenY = My / P = 28.03/3011.29 = 9 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(1,22) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(9,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #32 - 10 nos. (8042 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1608 0.00306 354.46 8.92 345.54 555.81 169 93.93
2 1608 0.00241 342.70 8.92 333.78 536.88 85 45.37
3 1608 0.00175 314.19 8.78 305.41 491.25 0 0.00
4 1608 0.00109 218.56 7.08 211.47 340.15 -85 -28.74
5 1608 0.00044 87.11 3.46 83.65 134.55 -169 -22.74
Total2058.63 87.82
xux = 450 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 450/450 = 0.360
Puc = 0.360 x 20.00 x 300 x 450
= 972.00 kN
Pux1 = Puc + Pus(Total)
= 972.00 + (2058.63)
= 3030.63 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 450/450
= 0.416
Muc = 972.00 x (0.5 x 450 - 0.416 x 450) = 36.74 kN-m
Mux1 = Muc + Mus(Total)
= 36.74 + (87.82)
= 124.56 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 4021 0.00274 351.41 8.92 342.49 1377.23 94 129.46
2 4021 0.00082 163.06 5.79 157.27 632.42 -94 -59.45
Total2009.66 70.01
xuy = 323 mm               Puc = C1.fck.B.D
ku = 1.078
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.884
C1 = 0.446 x (1 - C3/6) = 0.380
Puc = 0.380 x 20.00 x 450 x 300
= 1026.69 kN
Puy1 = Puc + Pus(Total)
= 1026.69 + (2009.66)
= 3036.34 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.438
Muc = 1026.69 x (0.5 x 300 - 0.438 x 300) = 19.02 kN-m
Muy1 = Muc + Mus(Total)
= 19.02 + (70.01)
= 89.03 kN-m
Pu/Puz = 0.826
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.826, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((65.50/124.56)2.000) + ((60.23/89.03)2.000)
= 0.734 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 32/4 = 8 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (8, 6) = 8 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 32 = 512 mm
               iii) 300 mm
               Required spacing = minimum of (300, 512, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG38 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #32 - 10 nos. (8042 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG39 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 439.13 kN
MomentX,(Mx) = 21.49 kN-m
MomentY,(My) = 20.80 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C470.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 9.090 kN.m
My_MinEccen = 8.783 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 21.489 kN.m
My = max(My,My_MinEccen) + MuaddY = 20.803 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 21.49/439.13 = 49 mm
Actual eccenY = My / P = 20.80/439.13 = 47 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(49,21) = 49 mm
eccenY = max(Actual eccenY,eccenYMin) = max(47,20) = 47 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(49 mm) > 0.05 x 300(15 mm)
and eccenY(47 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00275 351.52 8.92 342.60 77.50 179 13.87
2 226 0.00129 257.71 7.79 249.92 56.53 90 5.06
3 226 -0.00017 -34.43 0.00 -34.43 -7.79 0 0.00
4 226 -0.00163 -306.88 0.00 -306.88 -69.41 -90 6.21
5 226 -0.00309 -354.72 0.00 -354.72 -80.24 -179 14.36
Total-23.41 39.51
xux = 214 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 214/450 = 0.172
Puc = 0.172 x 20.00 x 300 x 450
= 463.22 kN
Pux1 = Puc + Pus(Total)
= 463.22 + (-23.41)
= 439.81 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 214/450
= 0.198
Muc = 463.22 x (0.5 x 450 - 0.198 x 450) = 62.90 kN-m
Mux1 = Muc + Mus(Total)
= 62.90 + (39.51)
= 102.41 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00235 340.43 8.92 331.51 187.46 104 19.50
2 565 -0.00287 -352.81 0.00 -352.81 -199.51 -104 20.75
Total-12.04 40.25
xuy = 139 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 139/300 = 0.167
Puc = 0.167 x 20.00 x 450 x 300
= 451.83 kN
Puy1 = Puc + Pus(Total)
= 451.83 + (-12.04)
= 439.79 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 139/300
= 0.193
Muc = 451.83 x (0.5 x 300 - 0.193 x 300) = 41.56 kN-m
Muy1 = Muc + Mus(Total)
= 41.56 + (40.25)
= 81.81 kN-m
Pu/Puz = 0.282
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.282, an = 1.137
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((21.49/102.41)1.137) + ((20.80/81.81)1.137)
= 0.380 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG39 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 920.41 kN
MomentX,(Mx) = 20.78 kN-m
MomentY,(My) = 18.93 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C470.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 19.052 kN.m
My_MinEccen = 18.408 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 20.780 kN.m
My = max(My,My_MinEccen) + MuaddY = 18.929 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 20.78/920.41 = 23 mm
Actual eccenY = My / P = 18.93/920.41 = 21 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(23,21) = 23 mm
eccenY = max(Actual eccenY,eccenYMin) = max(21,20) = 21 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(23 mm) > 0.05 x 300(15 mm)
and eccenY(21 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00303 354.14 8.92 345.22 78.09 179 13.98
2 226 0.00211 331.75 8.92 322.83 73.02 90 6.54
3 226 0.00119 238.14 7.46 230.68 52.18 0 0.00
4 226 0.00027 54.43 2.26 52.17 11.80 -90 -1.06
5 226 -0.00065 -129.29 0.00 -129.29 -29.24 -179 5.23
Total185.85 24.69
xux = 341 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 341/450 = 0.273
Puc = 0.273 x 20.00 x 300 x 450
= 736.59 kN
Pux1 = Puc + Pus(Total)
= 736.59 + (185.85)
= 922.44 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 341/450
= 0.315
Muc = 736.59 x (0.5 x 450 - 0.315 x 450) = 61.24 kN-m
Mux1 = Muc + Mus(Total)
= 61.24 + (24.69)
= 85.93 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00282 352.29 8.92 343.37 194.17 104 20.19
2 565 -0.00027 -54.84 0.00 -54.84 -31.01 -104 3.23
Total163.16 23.42
xuy = 236 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 236/300 = 0.283
Puc = 0.283 x 20.00 x 450 x 300
= 763.17 kN
Puy1 = Puc + Pus(Total)
= 763.17 + (163.16)
= 926.33 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 236/300
= 0.327
Muc = 763.17 x (0.5 x 300 - 0.327 x 300) = 39.69 kN-m
Muy1 = Muc + Mus(Total)
= 39.69 + (23.42)
= 63.11 kN-m
Pu/Puz = 0.591
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.591, an = 1.652
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((20.78/85.93)1.652) + ((18.93/63.11)1.652)
= 0.233 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG39 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1365.17 kN
MomentX,(Mx) = 14.27 kN-m
MomentY,(My) = 15.68 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C470.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 28.259 kN.m
My_MinEccen = 27.303 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 28.259 kN.m
My = max(My,My_MinEccen) + MuaddY = 27.303 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 14.27/1365.17 = 10 mm
Actual eccenY = My / P = 15.68/1365.17 = 11 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(10,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(11,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 6 nos. + #16 - 4 nos. (1483 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00307 354.50 8.92 345.58 138.96 177 24.60
2 226 0.00241 342.73 8.92 333.81 75.51 88 6.61
3 226 0.00176 315.00 8.79 306.20 69.26 0 0.00
4 226 0.00112 223.54 7.18 216.36 48.94 -88 -4.28
5 402 0.00046 91.54 3.62 87.92 35.35 -177 -6.26
Total368.03 20.66
xux = 464 mm               Puc = C1.fck.B.D
ku = 1.031
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.027
C1 = 0.446 x (1 - C3/6) = 0.370
Puc = 0.370 x 20.00 x 300 x 450
= 998.00 kN
Pux1 = Puc + Pus(Total)
= 998.00 + (368.03)
= 1366.03 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.426
Muc = 998.00 x (0.5 x 450 - 0.426 x 450) = 33.14 kN-m
Mux1 = Muc + Mus(Total)
= 33.14 + (20.66)
= 53.80 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 741 0.00285 352.56 8.92 343.64 254.78 102 25.99
2 741 0.00070 140.39 5.16 135.23 100.26 -102 -10.23
Total355.05 15.76
xuy = 319 mm               Puc = C1.fck.B.D
ku = 1.063
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.929
C1 = 0.446 x (1 - C3/6) = 0.377
Puc = 0.377 x 20.00 x 450 x 300
= 1017.83 kN
Puy1 = Puc + Pus(Total)
= 1017.83 + (355.05)
= 1372.87 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.435
Muc = 1017.83 x (0.5 x 300 - 0.435 x 300) = 19.97 kN-m
Muy1 = Muc + Mus(Total)
= 19.97 + (15.76)
= 35.73 kN-m
Pu/Puz = 0.821
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.821, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((28.26/53.80)2.000) + ((27.30/35.73)2.000)
= 0.860 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG39 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 6 nos. + #16 - 4 nos. (1483 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1748.02 kN
MomentX,(Mx) = 6.33 kN-m
MomentY,(My) = 10.27 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C470.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 36.184 kN.m
My_MinEccen = 34.960 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 36.184 kN.m
My = max(My,My_MinEccen) + MuaddY = 34.960 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 6.33/1748.02 = 4 mm
Actual eccenY = My / P = 10.27/1748.02 = 6 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(4,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(6,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 16 nos. (3217 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00311 354.90 8.92 345.98 139.13 177 24.63
2 402 0.00271 350.48 8.92 341.56 137.35 126 17.36
3 402 0.00230 338.85 8.92 329.93 132.67 76 10.06
4 402 0.00190 323.34 8.90 314.44 126.44 25 3.20
5 402 0.00149 293.49 8.34 285.15 114.66 -25 -2.90
6 402 0.00108 216.90 7.05 209.85 84.39 -76 -6.40
7 402 0.00068 135.70 5.03 130.67 52.55 -126 -6.64
8 402 0.00027 54.49 2.26 52.23 21.00 -177 -3.72
Total808.20 35.59
xux = 436 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 436/450 = 0.349
Puc = 0.349 x 20.00 x 300 x 450
= 941.63 kN
Pux1 = Puc + Pus(Total)
= 941.63 + (808.20)
= 1749.82 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 436/450
= 0.403
Muc = 941.63 x (0.5 x 450 - 0.403 x 450) = 41.10 kN-m
Mux1 = Muc + Mus(Total)
= 41.10 + (35.59)
= 76.69 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1608 0.00288 352.85 8.92 343.93 553.21 102 56.43
2 1608 0.00065 130.43 4.87 125.56 201.96 -102 -20.60
Total755.17 35.83
xuy = 312 mm               Puc = C1.fck.B.D
ku = 1.039
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.001
C1 = 0.446 x (1 - C3/6) = 0.372
Puc = 0.372 x 20.00 x 450 x 300
= 1003.24 kN
Puy1 = Puc + Pus(Total)
= 1003.24 + (755.17)
= 1758.41 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.428
Muc = 1003.24 x (0.5 x 300 - 0.428 x 300) = 21.53 kN-m
Muy1 = Muc + Mus(Total)
= 21.53 + (35.83)
= 57.36 kN-m
Pu/Puz = 0.799
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.799, an = 1.999
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((36.18/76.69)1.999) + ((34.96/57.36)1.999)
= 0.595 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG39 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 16 nos. (3217 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 2006.86 kN
MomentX,(Mx) = 2.12 kN-m
MomentY,(My) = 3.64 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C470.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 43.649 kN.m
My_MinEccen = 40.137 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 43.649 kN.m
My = max(My,My_MinEccen) + MuaddY = 40.137 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 2.12/2006.86 = 1 mm
Actual eccenY = My / P = 3.64/2006.86 = 2 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(1,22) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(2,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 14 nos. (4398 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00309 354.70 8.92 345.78 217.26 175 38.02
2 628 0.00262 348.10 8.92 339.18 213.11 117 24.86
3 628 0.00214 332.88 8.92 323.96 203.55 58 11.87
4 628 0.00166 308.82 8.67 300.15 188.59 0 0.00
5 628 0.00119 237.58 7.45 230.13 144.60 -58 -8.43
6 628 0.00071 142.38 5.22 137.16 86.18 -117 -10.05
7 628 0.00024 47.18 1.98 45.20 28.40 -175 -4.97
Total1081.69 51.30
xux = 429 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 429/450 = 0.343
Puc = 0.343 x 20.00 x 300 x 450
= 926.44 kN
Pux1 = Puc + Pus(Total)
= 926.44 + (1081.69)
= 2008.13 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 429/450
= 0.397
Muc = 926.44 x (0.5 x 450 - 0.397 x 450) = 43.15 kN-m
Mux1 = Muc + Mus(Total)
= 43.15 + (51.30)
= 94.45 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 2199 0.00288 352.85 8.92 343.93 756.35 100 75.64
2 2199 0.00064 127.83 4.79 123.04 270.58 -100 -27.06
Total1026.93 48.58
xuy = 307 mm               Puc = C1.fck.B.D
ku = 1.023
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.055
C1 = 0.446 x (1 - C3/6) = 0.368
Puc = 0.368 x 20.00 x 450 x 300
= 992.55 kN
Puy1 = Puc + Pus(Total)
= 992.55 + (1026.93)
= 2019.48 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.424
Muc = 992.55 x (0.5 x 300 - 0.424 x 300) = 22.68 kN-m
Muy1 = Muc + Mus(Total)
= 22.68 + (48.58)
= 71.25 kN-m
Pu/Puz = 0.789
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.789, an = 1.981
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((43.65/94.45)1.981) + ((40.14/71.25)1.981)
= 0.537 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 20 = 320 mm
               iii) 300 mm
               Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG39 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 14 nos. (4398 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG40 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 383.87 kN
MomentX,(Mx) = 18.72 kN-m
MomentY,(My) = 0.03 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C480.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 7.946 kN.m
My_MinEccen = 7.677 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 18.723 kN.m
My = max(My,My_MinEccen) + MuaddY = 7.677 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 18.72/383.87 = 49 mm
Actual eccenY = My / P = 0.03/383.87 = 0 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(49,21) = 49 mm
eccenY = max(Actual eccenY,eccenYMin) = max(0,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(49 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00270 350.17 8.92 341.25 77.19 179 13.82
2 226 0.00113 226.67 7.25 219.43 49.63 90 4.44
3 226 -0.00043 -85.96 0.00 -85.96 -19.44 0 0.00
4 226 -0.00199 -327.48 0.00 -327.48 -74.07 -90 6.63
5 226 -0.00356 -358.77 0.00 -358.77 -81.15 -179 14.53
Total-47.85 39.41
xux = 200 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 200/450 = 0.160
Puc = 0.160 x 20.00 x 300 x 450
= 432.84 kN
Pux1 = Puc + Pus(Total)
= 432.84 + (-47.85)
= 385.00 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 200/450
= 0.185
Muc = 432.84 x (0.5 x 450 - 0.185 x 450) = 61.31 kN-m
Mux1 = Muc + Mus(Total)
= 61.31 + (39.41)
= 100.72 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00221 335.45 8.92 326.53 184.65 104 19.20
2 565 -0.00362 -359.35 0.00 -359.35 -203.21 -104 21.13
Total-18.56 40.34
xuy = 125 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 125/300 = 0.150
Puc = 0.150 x 20.00 x 450 x 300
= 404.37 kN
Puy1 = Puc + Pus(Total)
= 404.37 + (-18.56)
= 385.81 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 125/300
= 0.173
Muc = 404.37 x (0.5 x 300 - 0.173 x 300) = 39.66 kN-m
Muy1 = Muc + Mus(Total)
= 39.66 + (40.34)
= 80.00 kN-m
Pu/Puz = 0.247
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.247, an = 1.078
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((18.72/100.72)1.078) + ((7.68/80.00)1.078)
= 0.243 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG40 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 797.55 kN
MomentX,(Mx) = 19.68 kN-m
MomentY,(My) = 0.33 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C480.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 16.509 kN.m
My_MinEccen = 15.951 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 19.684 kN.m
My = max(My,My_MinEccen) + MuaddY = 15.951 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 19.68/797.55 = 25 mm
Actual eccenY = My / P = 0.33/797.55 = 0 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(25,21) = 25 mm
eccenY = max(Actual eccenY,eccenYMin) = max(0,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(25 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00298 353.70 8.92 344.78 77.99 179 13.96
2 226 0.00196 326.21 8.92 317.29 71.77 90 6.42
3 226 0.00094 188.00 6.41 181.59 41.07 0 0.00
4 226 -0.00008 -15.66 0.00 -15.66 -3.54 -90 0.32
5 226 -0.00110 -219.32 0.00 -219.32 -49.61 -179 8.88
Total137.68 29.58
xux = 308 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 308/450 = 0.246
Puc = 0.246 x 20.00 x 300 x 450
= 664.45 kN
Pux1 = Puc + Pus(Total)
= 664.45 + (137.68)
= 802.13 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 308/450
= 0.284
Muc = 664.45 x (0.5 x 450 - 0.284 x 450) = 64.47 kN-m
Mux1 = Muc + Mus(Total)
= 64.47 + (29.58)
= 94.05 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00274 351.31 8.92 342.39 193.62 104 20.14
2 565 -0.00069 -138.25 0.00 -138.25 -78.18 -104 8.13
Total115.44 28.27
xuy = 212 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 212/300 = 0.255
Puc = 0.255 x 20.00 x 450 x 300
= 687.23 kN
Puy1 = Puc + Pus(Total)
= 687.23 + (115.44)
= 802.67 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 212/300
= 0.294
Muc = 687.23 x (0.5 x 300 - 0.294 x 300) = 42.45 kN-m
Muy1 = Muc + Mus(Total)
= 42.45 + (28.27)
= 70.71 kN-m
Pu/Puz = 0.512
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.512, an = 1.520
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((19.68/94.05)1.520) + ((15.95/70.71)1.520)
= 0.197 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG40 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1204.60 kN
MomentX,(Mx) = 12.03 kN-m
MomentY,(My) = 0.55 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C480.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 24.935 kN.m
My_MinEccen = 24.092 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 24.935 kN.m
My = max(My,My_MinEccen) + MuaddY = 24.092 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 12.03/1204.60 = 10 mm
Actual eccenY = My / P = 0.55/1204.60 = 0 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(10,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(0,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00313 355.02 8.92 346.10 78.29 179 14.01
2 226 0.00240 342.55 8.92 333.63 75.47 90 6.75
3 226 0.00168 309.75 8.69 301.06 68.10 0 0.00
4 226 0.00095 190.89 6.48 184.41 41.71 -90 -3.73
5 226 0.00023 46.01 1.93 44.08 9.97 -179 -1.78
Total273.53 15.25
xux = 432 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 432/450 = 0.346
Puc = 0.346 x 20.00 x 300 x 450
= 934.03 kN
Pux1 = Puc + Pus(Total)
= 934.03 + (273.53)
= 1207.56 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 432/450
= 0.400
Muc = 934.03 x (0.5 x 450 - 0.400 x 450) = 42.14 kN-m
Mux1 = Muc + Mus(Total)
= 42.14 + (15.25)
= 57.39 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00295 353.50 8.92 344.58 194.86 104 20.27
2 565 0.00049 97.93 3.83 94.09 53.21 -104 -5.53
Total248.07 14.73
xuy = 295 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 295/300 = 0.354
Puc = 0.354 x 20.00 x 450 x 300
= 956.81 kN
Puy1 = Puc + Pus(Total)
= 956.81 + (248.07)
= 1204.88 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 295/300
= 0.410
Muc = 956.81 x (0.5 x 300 - 0.410 x 300) = 25.98 kN-m
Muy1 = Muc + Mus(Total)
= 25.98 + (14.73)
= 40.71 kN-m
Pu/Puz = 0.774
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.774, an = 1.956
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((24.94/57.39)1.956) + ((24.09/40.71)1.956)
= 0.554 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG40 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1604.70 kN
MomentX,(Mx) = 2.81 kN-m
MomentY,(My) = 0.76 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C480.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 33.217 kN.m
My_MinEccen = 32.094 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 33.217 kN.m
My = max(My,My_MinEccen) + MuaddY = 32.094 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 2.81/1604.70 = 2 mm
Actual eccenY = My / P = 0.76/1604.70 = 0 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(2,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(0,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 12 nos. (2413 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00310 354.75 8.92 345.83 139.06 177 24.61
2 402 0.00256 346.67 8.92 337.75 135.82 106 14.42
3 402 0.00202 328.64 8.92 319.72 128.57 35 4.55
4 402 0.00149 293.31 8.34 284.97 114.59 -35 -4.06
5 402 0.00095 190.53 6.47 184.05 74.01 -106 -7.86
6 402 0.00042 83.33 3.33 80.00 32.17 -177 -5.69
Total624.23 25.98
xux = 457 mm               Puc = C1.fck.B.D
ku = 1.016
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.083
C1 = 0.446 x (1 - C3/6) = 0.366
Puc = 0.366 x 20.00 x 300 x 450
= 986.88 kN
Pux1 = Puc + Pus(Total)
= 986.88 + (624.23)
= 1611.10 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.421
Muc = 986.88 x (0.5 x 450 - 0.421 x 450) = 34.93 kN-m
Mux1 = Muc + Mus(Total)
= 34.93 + (25.98)
= 60.91 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1206 0.00283 352.39 8.92 343.47 414.35 102 42.26
2 1206 0.00073 146.64 5.34 141.30 170.46 -102 -17.39
Total584.81 24.88
xuy = 323 mm               Puc = C1.fck.B.D
ku = 1.078
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.884
C1 = 0.446 x (1 - C3/6) = 0.380
Puc = 0.380 x 20.00 x 450 x 300
= 1026.69 kN
Puy1 = Puc + Pus(Total)
= 1026.69 + (584.81)
= 1611.49 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.438
Muc = 1026.69 x (0.5 x 300 - 0.438 x 300) = 19.02 kN-m
Muy1 = Muc + Mus(Total)
= 19.02 + (24.88)
= 43.90 kN-m
Pu/Puz = 0.825
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.825, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((33.22/60.91)2.000) + ((32.09/43.90)2.000)
= 0.832 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG40 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 12 nos. (2413 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 2006.27 kN
MomentX,(Mx) = 2.98 kN-m
MomentY,(My) = 0.91 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C480.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 43.636 kN.m
My_MinEccen = 40.125 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 43.636 kN.m
My = max(My,My_MinEccen) + MuaddY = 40.125 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 2.98/2006.27 = 1 mm
Actual eccenY = My / P = 0.91/2006.27 = 0 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(1,22) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(0,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 14 nos. (4398 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00309 354.70 8.92 345.78 217.26 175 38.02
2 628 0.00262 348.10 8.92 339.18 213.11 117 24.86
3 628 0.00214 332.88 8.92 323.96 203.55 58 11.87
4 628 0.00166 308.82 8.67 300.15 188.59 0 0.00
5 628 0.00119 237.58 7.45 230.13 144.60 -58 -8.43
6 628 0.00071 142.38 5.22 137.16 86.18 -117 -10.05
7 628 0.00024 47.18 1.98 45.20 28.40 -175 -4.97
Total1081.69 51.30
xux = 429 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 429/450 = 0.343
Puc = 0.343 x 20.00 x 300 x 450
= 926.44 kN
Pux1 = Puc + Pus(Total)
= 926.44 + (1081.69)
= 2008.13 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 429/450
= 0.397
Muc = 926.44 x (0.5 x 450 - 0.397 x 450) = 43.15 kN-m
Mux1 = Muc + Mus(Total)
= 43.15 + (51.30)
= 94.45 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 2199 0.00289 352.96 8.92 344.04 756.58 100 75.66
2 2199 0.00062 124.21 4.68 119.53 262.86 -100 -26.29
Total1019.43 49.37
xuy = 305 mm               Puc = C1.fck.B.D
ku = 1.016
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.083
C1 = 0.446 x (1 - C3/6) = 0.366
Puc = 0.366 x 20.00 x 450 x 300
= 986.88 kN
Puy1 = Puc + Pus(Total)
= 986.88 + (1019.43)
= 2006.31 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.421
Muc = 986.88 x (0.5 x 300 - 0.421 x 300) = 23.28 kN-m
Muy1 = Muc + Mus(Total)
= 23.28 + (49.37)
= 72.66 kN-m
Pu/Puz = 0.789
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.789, an = 1.981
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((43.64/94.45)1.981) + ((40.13/72.66)1.981)
= 0.525 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 20 = 320 mm
               iii) 300 mm
               Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG40 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 14 nos. (4398 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG41 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 440.30 kN
MomentX,(Mx) = 16.50 kN-m
MomentY,(My) = 18.10 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C490.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 9.114 kN.m
My_MinEccen = 8.806 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 16.496 kN.m
My = max(My,My_MinEccen) + MuaddY = 18.096 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 16.50/440.30 = 37 mm
Actual eccenY = My / P = 18.10/440.30 = 41 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(37,21) = 37 mm
eccenY = max(Actual eccenY,eccenYMin) = max(41,20) = 41 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(37 mm) > 0.05 x 300(15 mm)
and eccenY(41 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00275 351.60 8.92 342.68 77.51 179 13.87
2 226 0.00130 259.52 7.82 251.70 56.93 90 5.10
3 226 -0.00016 -31.43 0.00 -31.43 -7.11 0 0.00
4 226 -0.00161 -304.98 0.00 -304.98 -68.99 -90 6.17
5 226 -0.00307 -354.48 0.00 -354.48 -80.18 -179 14.35
Total-21.83 39.50
xux = 215 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 215/450 = 0.172
Puc = 0.172 x 20.00 x 300 x 450
= 465.12 kN
Pux1 = Puc + Pus(Total)
= 465.12 + (-21.83)
= 443.29 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 215/450
= 0.199
Muc = 465.12 x (0.5 x 450 - 0.199 x 450) = 62.99 kN-m
Mux1 = Muc + Mus(Total)
= 62.99 + (39.50)
= 102.48 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00236 340.78 8.92 331.86 187.66 104 19.52
2 565 -0.00282 -352.34 0.00 -352.34 -199.24 -104 20.72
Total-11.58 40.24
xuy = 141 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 141/300 = 0.169
Puc = 0.169 x 20.00 x 450 x 300
= 455.63 kN
Puy1 = Puc + Pus(Total)
= 455.63 + (-11.58)
= 444.05 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 141/300
= 0.195
Muc = 455.63 x (0.5 x 300 - 0.195 x 300) = 41.69 kN-m
Muy1 = Muc + Mus(Total)
= 41.69 + (40.24)
= 81.93 kN-m
Pu/Puz = 0.283
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.283, an = 1.138
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((16.50/102.48)1.138) + ((18.10/81.93)1.138)
= 0.304 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG41 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 913.17 kN
MomentX,(Mx) = 12.35 kN-m
MomentY,(My) = 15.85 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C490.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 18.903 kN.m
My_MinEccen = 18.263 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 18.903 kN.m
My = max(My,My_MinEccen) + MuaddY = 18.263 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 12.35/913.17 = 14 mm
Actual eccenY = My / P = 15.85/913.17 = 17 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(14,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(17,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00303 354.12 8.92 345.20 78.08 179 13.98
2 226 0.00210 331.49 8.92 322.57 72.96 90 6.53
3 226 0.00118 235.75 7.42 228.34 51.65 0 0.00
4 226 0.00026 51.08 2.13 48.95 11.07 -90 -0.99
5 226 -0.00067 -133.58 0.00 -133.58 -30.22 -179 5.41
Total183.55 24.92
xux = 339 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 339/450 = 0.271
Puc = 0.271 x 20.00 x 300 x 450
= 732.80 kN
Pux1 = Puc + Pus(Total)
= 732.80 + (183.55)
= 916.35 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 339/450
= 0.314
Muc = 732.80 x (0.5 x 450 - 0.314 x 450) = 61.46 kN-m
Mux1 = Muc + Mus(Total)
= 61.46 + (24.92)
= 86.38 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00281 352.26 8.92 343.34 194.16 104 20.19
2 565 -0.00029 -58.61 0.00 -58.61 -33.15 -104 3.45
Total161.01 23.64
xuy = 234 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 234/300 = 0.281
Puc = 0.281 x 20.00 x 450 x 300
= 759.38 kN
Puy1 = Puc + Pus(Total)
= 759.38 + (161.01)
= 920.39 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 234/300
= 0.325
Muc = 759.38 x (0.5 x 300 - 0.325 x 300) = 39.87 kN-m
Muy1 = Muc + Mus(Total)
= 39.87 + (23.64)
= 63.51 kN-m
Pu/Puz = 0.587
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.587, an = 1.644
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((18.90/86.38)1.644) + ((18.26/63.51)1.644)
= 0.211 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG41 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1356.81 kN
MomentX,(Mx) = 7.92 kN-m
MomentY,(My) = 12.43 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C490.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 28.086 kN.m
My_MinEccen = 27.136 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 28.086 kN.m
My = max(My,My_MinEccen) + MuaddY = 27.136 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 7.92/1356.81 = 6 mm
Actual eccenY = My / P = 12.43/1356.81 = 9 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(6,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(9,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 2 nos. + #16 - 6 nos. (1433 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00304 354.26 8.92 345.34 138.87 177 24.58
2 226 0.00220 335.01 8.92 326.09 73.76 60 4.40
3 402 0.00135 270.89 7.99 262.90 105.72 -58 -6.10
4 402 0.00050 99.33 3.88 95.45 38.38 -177 -6.79
Total356.73 16.09
xux = 471 mm               Puc = C1.fck.B.D
ku = 1.047
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.976
C1 = 0.446 x (1 - C3/6) = 0.373
Puc = 0.373 x 20.00 x 300 x 450
= 1008.29 kN
Pux1 = Puc + Pus(Total)
= 1008.29 + (356.73)
= 1365.02 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.431
Muc = 1008.29 x (0.5 x 450 - 0.431 x 450) = 31.49 kN-m
Mux1 = Muc + Mus(Total)
= 31.49 + (16.09)
= 47.58 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 716 0.00285 352.56 8.92 343.64 246.15 102 25.11
2 716 0.00070 140.39 5.16 135.23 96.86 -102 -9.88
Total343.01 15.23
xuy = 319 mm               Puc = C1.fck.B.D
ku = 1.063
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.929
C1 = 0.446 x (1 - C3/6) = 0.377
Puc = 0.377 x 20.00 x 450 x 300
= 1017.83 kN
Puy1 = Puc + Pus(Total)
= 1017.83 + (343.01)
= 1360.84 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.435
Muc = 1017.83 x (0.5 x 300 - 0.435 x 300) = 19.97 kN-m
Muy1 = Muc + Mus(Total)
= 19.97 + (15.23)
= 35.20 kN-m
Pu/Puz = 0.823
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.823, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((28.09/47.58)2.000) + ((27.14/35.20)2.000)
= 0.943 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG41 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 2 nos. + #16 - 6 nos. (1433 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1750.12 kN
MomentX,(Mx) = 4.93 kN-m
MomentY,(My) = 7.11 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C490.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 36.227 kN.m
My_MinEccen = 35.002 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 36.227 kN.m
My = max(My,My_MinEccen) + MuaddY = 35.002 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 4.93/1750.12 = 3 mm
Actual eccenY = My / P = 7.11/1750.12 = 4 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(3,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(4,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 16 nos. (3217 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00312 354.93 8.92 346.01 139.14 177 24.63
2 402 0.00271 350.64 8.92 341.72 137.41 126 17.37
3 402 0.00231 339.21 8.92 330.29 132.82 76 10.08
4 402 0.00191 324.14 8.90 315.24 126.76 25 3.21
5 402 0.00151 295.01 8.38 286.63 115.26 -25 -2.91
6 402 0.00110 220.77 7.13 213.64 85.91 -76 -6.52
7 402 0.00070 140.21 5.16 135.06 54.31 -126 -6.87
8 402 0.00030 59.66 2.46 57.20 23.00 -177 -4.07
Total814.61 34.91
xux = 439 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 439/450 = 0.352
Puc = 0.352 x 20.00 x 300 x 450
= 949.22 kN
Pux1 = Puc + Pus(Total)
= 949.22 + (814.61)
= 1763.83 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 439/450
= 0.406
Muc = 949.22 x (0.5 x 450 - 0.406 x 450) = 40.05 kN-m
Mux1 = Muc + Mus(Total)
= 40.05 + (34.91)
= 74.96 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1608 0.00288 352.85 8.92 343.93 553.21 102 56.43
2 1608 0.00065 130.43 4.87 125.56 201.96 -102 -20.60
Total755.17 35.83
xuy = 312 mm               Puc = C1.fck.B.D
ku = 1.039
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.001
C1 = 0.446 x (1 - C3/6) = 0.372
Puc = 0.372 x 20.00 x 450 x 300
= 1003.24 kN
Puy1 = Puc + Pus(Total)
= 1003.24 + (755.17)
= 1758.41 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.428
Muc = 1003.24 x (0.5 x 300 - 0.428 x 300) = 21.53 kN-m
Muy1 = Muc + Mus(Total)
= 21.53 + (35.83)
= 57.36 kN-m
Pu/Puz = 0.800
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.800, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((36.23/74.96)2.000) + ((35.00/57.36)2.000)
= 0.606 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG41 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 16 nos. (3217 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 2040.87 kN
MomentX,(Mx) = 1.74 kN-m
MomentY,(My) = 1.20 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C490.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 44.389 kN.m
My_MinEccen = 40.817 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 44.389 kN.m
My = max(My,My_MinEccen) + MuaddY = 40.817 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 1.74/2040.87 = 1 mm
Actual eccenY = My / P = 1.20/2040.87 = 1 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(1,22) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(1,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 14 nos. (4398 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00310 354.76 8.92 345.84 217.30 175 38.03
2 628 0.00263 348.46 8.92 339.54 213.34 117 24.89
3 628 0.00216 333.69 8.92 324.77 204.06 58 11.90
4 628 0.00169 310.67 8.71 301.96 189.72 0 0.00
5 628 0.00123 245.04 7.58 237.46 149.20 -58 -8.70
6 628 0.00076 151.37 5.47 145.90 91.67 -117 -10.70
7 628 0.00029 57.71 2.39 55.32 34.76 -175 -6.08
Total1100.05 49.34
xux = 436 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 436/450 = 0.349
Puc = 0.349 x 20.00 x 300 x 450
= 941.63 kN
Pux1 = Puc + Pus(Total)
= 941.63 + (1100.05)
= 2041.68 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 436/450
= 0.403
Muc = 941.63 x (0.5 x 450 - 0.403 x 450) = 41.10 kN-m
Mux1 = Muc + Mus(Total)
= 41.10 + (49.34)
= 90.44 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 2199 0.00285 352.56 8.92 343.64 755.71 100 75.57
2 2199 0.00069 138.15 5.10 133.05 292.59 -100 -29.26
Total1048.30 46.31
xuy = 314 mm               Puc = C1.fck.B.D
ku = 1.047
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.976
C1 = 0.446 x (1 - C3/6) = 0.373
Puc = 0.373 x 20.00 x 450 x 300
= 1008.29 kN
Puy1 = Puc + Pus(Total)
= 1008.29 + (1048.30)
= 2056.59 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.431
Muc = 1008.29 x (0.5 x 300 - 0.431 x 300) = 20.99 kN-m
Muy1 = Muc + Mus(Total)
= 20.99 + (46.31)
= 67.30 kN-m
Pu/Puz = 0.802
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.802, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((44.39/90.44)2.000) + ((40.82/67.30)2.000)
= 0.609 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 20 = 320 mm
               iii) 300 mm
               Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG41 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 14 nos. (4398 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG42 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 553.99 kN
MomentX,(Mx) = 11.88 kN-m
MomentY,(My) = 53.59 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C500.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 11.468 kN.m
My_MinEccen = 11.080 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 11.877 kN.m
My = max(My,My_MinEccen) + MuaddY = 53.593 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 11.88/553.99 = 21 mm
Actual eccenY = My / P = 53.59/553.99 = 97 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(21,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(97,20) = 97 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(97 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00284 352.47 8.92 343.55 77.71 179 13.91
2 226 0.00154 298.64 8.46 290.19 65.64 90 5.87
3 226 0.00025 50.72 2.12 48.61 10.99 0 0.00
4 226 -0.00104 -207.54 0.00 -207.54 -46.95 -90 4.20
5 226 -0.00233 -339.83 0.00 -339.83 -76.87 -179 13.76
Total30.53 37.75
xux = 243 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 243/450 = 0.194
Puc = 0.194 x 20.00 x 300 x 450
= 523.97 kN
Pux1 = Puc + Pus(Total)
= 523.97 + (30.53)
= 554.50 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 243/450
= 0.224
Muc = 523.97 x (0.5 x 450 - 0.224 x 450) = 65.02 kN-m
Mux1 = Muc + Mus(Total)
= 65.02 + (37.75)
= 102.76 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00255 346.30 8.92 337.38 190.78 104 19.84
2 565 -0.00177 -315.32 0.00 -315.32 -178.31 -104 18.54
Total12.47 38.39
xuy = 169 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 169/300 = 0.203
Puc = 0.203 x 20.00 x 450 x 300
= 546.75 kN
Puy1 = Puc + Pus(Total)
= 546.75 + (12.47)
= 559.22 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 169/300
= 0.234
Muc = 546.75 x (0.5 x 300 - 0.234 x 300) = 43.63 kN-m
Muy1 = Muc + Mus(Total)
= 43.63 + (38.39)
= 82.02 kN-m
Pu/Puz = 0.356
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.356, an = 1.260
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((11.88/102.76)1.260) + ((53.59/82.02)1.260)
= 0.651 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG42 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1085.60 kN
MomentX,(Mx) = 8.49 kN-m
MomentY,(My) = 49.59 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C500.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 22.472 kN.m
My_MinEccen = 21.712 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 22.472 kN.m
My = max(My,My_MinEccen) + MuaddY = 49.594 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 8.49/1085.60 = 8 mm
Actual eccenY = My / P = 49.59/1085.60 = 46 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(8,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(46,20) = 46 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(46 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 4 nos. + #16 - 4 nos. (1257 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00307 354.51 8.92 345.59 138.97 177 24.60
2 226 0.00201 327.93 8.92 319.01 72.16 58 4.21
3 226 0.00096 191.76 6.50 185.26 41.90 -58 -2.44
4 402 -0.00011 -21.11 0.00 -21.11 -8.49 -177 1.50
Total244.54 27.86
xux = 390 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 390/450 = 0.312
Puc = 0.312 x 20.00 x 300 x 450
= 842.91 kN
Pux1 = Puc + Pus(Total)
= 842.91 + (244.54)
= 1087.45 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 390/450
= 0.361
Muc = 842.91 x (0.5 x 450 - 0.361 x 450) = 52.82 kN-m
Mux1 = Muc + Mus(Total)
= 52.82 + (27.86)
= 80.68 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00287 352.72 8.92 343.80 216.02 102 22.03
2 628 0.00017 33.95 1.45 32.50 20.42 -102 -2.08
Total236.44 19.95
xuy = 265 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 265/300 = 0.318
Puc = 0.318 x 20.00 x 450 x 300
= 858.09 kN
Puy1 = Puc + Pus(Total)
= 858.09 + (236.44)
= 1094.53 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 265/300
= 0.367
Muc = 858.09 x (0.5 x 300 - 0.367 x 300) = 34.17 kN-m
Muy1 = Muc + Mus(Total)
= 34.17 + (19.95)
= 54.12 kN-m
Pu/Puz = 0.681
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.681, an = 1.801
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((22.47/80.68)1.801) + ((49.59/54.12)1.801)
= 0.954 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG42 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 4 nos. + #16 - 4 nos. (1257 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1647.91 kN
MomentX,(Mx) = 5.09 kN-m
MomentY,(My) = 46.62 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C500.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 34.112 kN.m
My_MinEccen = 32.958 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 34.112 kN.m
My = max(My,My_MinEccen) + MuaddY = 46.624 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 5.09/1647.91 = 3 mm
Actual eccenY = My / P = 46.62/1647.91 = 28 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(3,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(28,20) = 28 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(28 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 14 nos. (2815 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00312 354.93 8.92 346.01 139.14 177 24.63
2 402 0.00265 348.91 8.92 339.99 136.72 118 16.13
3 402 0.00218 334.27 8.92 325.35 130.83 59 7.72
4 402 0.00171 311.57 8.73 302.84 121.78 0 0.00
5 402 0.00124 247.62 7.63 239.99 96.51 -59 -5.69
6 402 0.00077 153.64 5.54 148.10 59.56 -118 -7.03
7 402 0.00030 59.66 2.46 57.20 23.00 -177 -4.07
Total707.53 31.69
xux = 439 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 439/450 = 0.352
Puc = 0.352 x 20.00 x 300 x 450
= 949.22 kN
Pux1 = Puc + Pus(Total)
= 949.22 + (707.53)
= 1656.75 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 439/450
= 0.406
Muc = 949.22 x (0.5 x 450 - 0.406 x 450) = 40.05 kN-m
Mux1 = Muc + Mus(Total)
= 40.05 + (31.69)
= 71.73 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1407 0.00289 352.95 8.92 344.03 484.20 102 49.39
2 1407 0.00063 126.93 4.76 122.17 171.95 -102 -17.54
Total656.14 31.85
xuy = 309 mm               Puc = C1.fck.B.D
ku = 1.031
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.027
C1 = 0.446 x (1 - C3/6) = 0.370
Puc = 0.370 x 20.00 x 450 x 300
= 998.00 kN
Puy1 = Puc + Pus(Total)
= 998.00 + (656.14)
= 1654.14 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.426
Muc = 998.00 x (0.5 x 300 - 0.426 x 300) = 22.09 kN-m
Muy1 = Muc + Mus(Total)
= 22.09 + (31.85)
= 53.94 kN-m
Pu/Puz = 0.798
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.798, an = 1.996
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((34.11/71.73)1.996) + ((46.62/53.94)1.996)
= 0.974 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG42 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 14 nos. (2815 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 2260.82 kN
MomentX,(Mx) = 2.99 kN-m
MomentY,(My) = 41.79 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C500.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 46.799 kN.m
My_MinEccen = 45.216 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 46.799 kN.m
My = max(My,My_MinEccen) + MuaddY = 45.216 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 2.99/2260.82 = 1 mm
Actual eccenY = My / P = 41.79/2260.82 = 18 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(1,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(18,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 16 nos. (5027 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00311 354.87 8.92 345.95 217.37 175 38.04
2 628 0.00272 350.83 8.92 341.91 214.83 125 26.85
3 628 0.00233 339.98 8.92 331.06 208.01 75 15.60
4 628 0.00194 325.70 8.91 316.78 199.04 25 4.98
5 628 0.00156 299.65 8.48 291.17 182.95 -25 -4.57
6 628 0.00117 233.33 7.37 225.96 141.98 -75 -10.65
7 628 0.00078 155.56 5.59 149.97 94.23 -125 -11.78
8 628 0.00039 77.78 3.13 74.65 46.90 -175 -8.21
Total1305.30 50.26
xux = 450 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 450/450 = 0.360
Puc = 0.360 x 20.00 x 300 x 450
= 972.00 kN
Pux1 = Puc + Pus(Total)
= 972.00 + (1305.30)
= 2277.30 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 450/450
= 0.416
Muc = 972.00 x (0.5 x 450 - 0.416 x 450) = 36.74 kN-m
Mux1 = Muc + Mus(Total)
= 36.74 + (50.26)
= 87.00 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 2513 0.00279 352.04 8.92 343.12 862.36 100 86.24
2 2513 0.00078 156.60 5.62 150.98 379.46 -100 -37.95
Total1241.82 48.29
xuy = 328 mm               Puc = C1.fck.B.D
ku = 1.094
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.843
C1 = 0.446 x (1 - C3/6) = 0.383
Puc = 0.383 x 20.00 x 450 x 300
= 1034.93 kN
Puy1 = Puc + Pus(Total)
= 1034.93 + (1241.82)
= 2276.74 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.442
Muc = 1034.93 x (0.5 x 300 - 0.442 x 300) = 18.14 kN-m
Muy1 = Muc + Mus(Total)
= 18.14 + (48.29)
= 66.43 kN-m
Pu/Puz = 0.827
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.827, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((46.80/87.00)2.000) + ((45.22/66.43)2.000)
= 0.753 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 20 = 320 mm
               iii) 300 mm
               Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG42 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 16 nos. (5027 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 2972.33 kN
MomentX,(Mx) = 0.94 kN-m
MomentY,(My) = 24.65 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C500.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 64.648 kN.m
My_MinEccen = 59.447 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 64.648 kN.m
My = max(My,My_MinEccen) + MuaddY = 59.447 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 0.94/2972.33 = 0 mm
Actual eccenY = My / P = 24.65/2972.33 = 8 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(0,22) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(8,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #32 - 10 nos. (8042 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1608 0.00306 354.40 8.92 345.48 555.71 169 93.91
2 1608 0.00239 342.06 8.92 333.14 535.86 85 45.28
3 1608 0.00172 312.46 8.75 303.71 488.51 0 0.00
4 1608 0.00105 210.91 6.93 203.99 328.11 -85 -27.73
5 1608 0.00039 77.38 3.12 74.27 119.46 -169 -20.19
Total2027.64 91.28
xux = 443 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 443/450 = 0.354
Puc = 0.354 x 20.00 x 300 x 450
= 956.81 kN
Pux1 = Puc + Pus(Total)
= 956.81 + (2027.64)
= 2984.46 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 443/450
= 0.410
Muc = 956.81 x (0.5 x 450 - 0.410 x 450) = 38.97 kN-m
Mux1 = Muc + Mus(Total)
= 38.97 + (91.28)
= 130.25 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 4021 0.00278 352.00 8.92 343.08 1379.59 94 129.68
2 4021 0.00076 151.09 5.47 145.62 585.57 -94 -55.04
Total1965.16 74.64
xuy = 314 mm               Puc = C1.fck.B.D
ku = 1.047
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.976
C1 = 0.446 x (1 - C3/6) = 0.373
Puc = 0.373 x 20.00 x 450 x 300
= 1008.29 kN
Puy1 = Puc + Pus(Total)
= 1008.29 + (1965.16)
= 2973.45 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.431
Muc = 1008.29 x (0.5 x 300 - 0.431 x 300) = 20.99 kN-m
Muy1 = Muc + Mus(Total)
= 20.99 + (74.64)
= 95.63 kN-m
Pu/Puz = 0.815
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.815, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((64.65/130.25)2.000) + ((59.45/95.63)2.000)
= 0.633 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 32/4 = 8 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (8, 6) = 8 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 32 = 512 mm
               iii) 300 mm
               Required spacing = minimum of (300, 512, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG42 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #32 - 10 nos. (8042 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG43 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 429.31 kN
MomentX,(Mx) = 1.07 kN-m
MomentY,(My) = 90.20 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C510.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 8.887 kN.m
My_MinEccen = 8.586 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 8.887 kN.m
My = max(My,My_MinEccen) + MuaddY = 90.197 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 1.07/429.31 = 3 mm
Actual eccenY = My / P = 90.20/429.31 = 210 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(3,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(210,20) = 210 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(210 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 14 nos. (1583 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00275 351.52 8.92 342.60 77.50 179 13.87
2 226 0.00178 315.78 8.81 306.97 69.44 119 8.29
3 226 0.00080 160.33 5.72 154.62 34.97 60 2.09
4 226 -0.00017 -34.43 0.00 -34.43 -7.79 -0 0.00
5 226 -0.00115 -229.19 0.00 -229.19 -51.84 -60 3.09
6 226 -0.00212 -332.14 0.00 -332.14 -75.13 -119 8.97
7 226 -0.00309 -354.72 0.00 -354.72 -80.24 -179 14.36
Total-33.09 50.66
xux = 214 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 214/450 = 0.172
Puc = 0.172 x 20.00 x 300 x 450
= 463.22 kN
Pux1 = Puc + Pus(Total)
= 463.22 + (-33.09)
= 430.13 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 214/450
= 0.198
Muc = 463.22 x (0.5 x 450 - 0.198 x 450) = 62.90 kN-m
Mux1 = Muc + Mus(Total)
= 62.90 + (50.66)
= 113.56 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 792 0.00234 340.07 8.92 331.15 262.17 104 27.27
2 792 -0.00293 -353.28 0.00 -353.28 -279.68 -104 29.09
Total-17.52 56.35
xuy = 138 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 138/300 = 0.166
Puc = 0.166 x 20.00 x 450 x 300
= 448.03 kN
Puy1 = Puc + Pus(Total)
= 448.03 + (-17.52)
= 430.51 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 138/300
= 0.192
Muc = 448.03 x (0.5 x 300 - 0.192 x 300) = 41.43 kN-m
Muy1 = Muc + Mus(Total)
= 41.43 + (56.35)
= 97.78 kN-m
Pu/Puz = 0.253
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.253, an = 1.089
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((8.89/113.56)1.089) + ((90.20/97.78)1.089)
= 0.978 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG43 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 14 nos. (1583 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 875.61 kN
MomentX,(Mx) = 2.04 kN-m
MomentY,(My) = 82.74 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C510.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 18.125 kN.m
My_MinEccen = 17.512 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 18.125 kN.m
My = max(My,My_MinEccen) + MuaddY = 82.737 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 2.04/875.61 = 2 mm
Actual eccenY = My / P = 82.74/875.61 = 94 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(2,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(94,20) = 94 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(94 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. + #16 - 4 nos. (1935 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00295 353.44 8.92 344.52 138.54 177 24.52
2 226 0.00225 337.04 8.92 328.12 74.22 117 8.66
3 226 0.00158 302.13 8.53 293.60 66.41 58 3.87
4 226 0.00091 182.08 6.27 175.81 39.77 0 0.00
5 226 0.00024 47.81 2.00 45.80 10.36 -58 -0.60
6 226 -0.00043 -86.47 0.00 -86.47 -19.56 -117 2.28
7 402 -0.00113 -225.35 0.00 -225.35 -90.62 -177 16.04
Total219.12 54.77
xux = 304 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 304/450 = 0.243
Puc = 0.243 x 20.00 x 300 x 450
= 656.86 kN
Pux1 = Puc + Pus(Total)
= 656.86 + (219.12)
= 875.98 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 304/450
= 0.281
Muc = 656.86 x (0.5 x 450 - 0.281 x 450) = 64.70 kN-m
Mux1 = Muc + Mus(Total)
= 64.70 + (54.77)
= 119.47 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 968 0.00270 350.35 8.92 341.43 330.37 102 33.70
2 968 -0.00068 -136.27 0.00 -136.27 -131.85 -102 13.45
Total198.52 47.15
xuy = 211 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 211/300 = 0.253
Puc = 0.253 x 20.00 x 450 x 300
= 683.44 kN
Puy1 = Puc + Pus(Total)
= 683.44 + (198.52)
= 881.95 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 211/300
= 0.293
Muc = 683.44 x (0.5 x 300 - 0.293 x 300) = 42.54 kN-m
Muy1 = Muc + Mus(Total)
= 42.54 + (47.15)
= 89.69 kN-m
Pu/Puz = 0.486
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.486, an = 1.477
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((18.13/119.47)1.477) + ((82.74/89.69)1.477)
= 0.949 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG43 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. + #16 - 4 nos. (1935 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1320.01 kN
MomentX,(Mx) = 1.42 kN-m
MomentY,(My) = 80.69 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C510.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 27.324 kN.m
My_MinEccen = 26.400 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 27.324 kN.m
My = max(My,My_MinEccen) + MuaddY = 80.689 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 1.42/1320.01 = 1 mm
Actual eccenY = My / P = 80.69/1320.01 = 61 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(1,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(61,20) = 61 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(61 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 16 nos. (3217 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00301 354.03 8.92 345.11 138.78 177 24.56
2 402 0.00250 345.21 8.92 336.29 135.23 126 17.10
3 402 0.00199 327.47 8.92 318.55 128.10 76 9.72
4 402 0.00148 292.63 8.32 284.31 114.33 25 2.89
5 402 0.00097 194.06 6.56 187.51 75.40 -25 -1.91
6 402 0.00046 91.84 3.63 88.21 35.47 -76 -2.69
7 402 -0.00005 -10.39 0.00 -10.39 -4.18 -126 0.53
8 402 -0.00056 -112.62 0.00 -112.62 -45.29 -177 8.02
Total577.84 58.21
xux = 346 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 346/450 = 0.277
Puc = 0.277 x 20.00 x 300 x 450
= 747.98 kN
Pux1 = Puc + Pus(Total)
= 747.98 + (577.84)
= 1325.82 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 346/450
= 0.320
Muc = 747.98 x (0.5 x 450 - 0.320 x 450) = 60.54 kN-m
Mux1 = Muc + Mus(Total)
= 60.54 + (58.21)
= 118.76 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1608 0.00282 352.30 8.92 343.38 552.33 102 56.34
2 1608 -0.00008 -16.80 0.00 -16.80 -27.02 -102 2.76
Total525.31 59.09
xuy = 246 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 246/300 = 0.295
Puc = 0.295 x 20.00 x 450 x 300
= 797.34 kN
Puy1 = Puc + Pus(Total)
= 797.34 + (525.31)
= 1322.65 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 246/300
= 0.341
Muc = 797.34 x (0.5 x 300 - 0.341 x 300) = 37.97 kN-m
Muy1 = Muc + Mus(Total)
= 37.97 + (59.09)
= 97.07 kN-m
Pu/Puz = 0.603
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.603, an = 1.672
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((27.32/118.76)1.672) + ((80.69/97.07)1.672)
= 0.820 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG43 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 16 nos. (3217 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1761.29 kN
MomentX,(Mx) = 0.38 kN-m
MomentY,(My) = 77.31 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C510.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 36.459 kN.m
My_MinEccen = 35.226 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 36.459 kN.m
My = max(My,My_MinEccen) + MuaddY = 77.314 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 0.38/1761.29 = 0 mm
Actual eccenY = My / P = 77.31/1761.29 = 44 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(0,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(44,20) = 44 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(44 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 14 nos. (4398 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00304 354.28 8.92 345.36 217.00 175 37.97
2 628 0.00251 345.39 8.92 336.47 211.41 117 24.66
3 628 0.00198 326.92 8.92 318.00 199.81 58 11.66
4 628 0.00144 289.17 8.23 280.94 176.52 0 0.00
5 628 0.00091 182.43 6.28 176.15 110.68 -58 -6.46
6 628 0.00038 75.87 3.06 72.81 45.75 -117 -5.34
7 628 -0.00015 -30.68 0.00 -30.68 -19.28 -175 3.37
Total941.88 65.87
xux = 383 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 383/450 = 0.307
Puc = 0.307 x 20.00 x 300 x 450
= 827.72 kN
Pux1 = Puc + Pus(Total)
= 827.72 + (941.88)
= 1769.60 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 383/450
= 0.354
Muc = 827.72 x (0.5 x 450 - 0.354 x 450) = 54.29 kN-m
Mux1 = Muc + Mus(Total)
= 54.29 + (65.87)
= 120.16 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 2199 0.00286 352.69 8.92 343.77 755.99 100 75.60
2 2199 0.00031 61.82 2.54 59.28 130.36 -100 -13.04
Total886.35 62.56
xuy = 274 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 274/300 = 0.329
Puc = 0.329 x 20.00 x 450 x 300
= 888.47 kN
Puy1 = Puc + Pus(Total)
= 888.47 + (886.35)
= 1774.82 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 274/300
= 0.380
Muc = 888.47 x (0.5 x 300 - 0.380 x 300) = 31.92 kN-m
Muy1 = Muc + Mus(Total)
= 31.92 + (62.56)
= 94.48 kN-m
Pu/Puz = 0.692
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.692, an = 1.820
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((36.46/120.16)1.820) + ((77.31/94.48)1.820)
= 0.808 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 20 = 320 mm
               iii) 300 mm
               Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG43 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 14 nos. (4398 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 2192.18 kN
MomentX,(Mx) = 0.95 kN-m
MomentY,(My) = 51.03 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C510.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 47.680 kN.m
My_MinEccen = 43.844 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 47.680 kN.m
My = max(My,My_MinEccen) + MuaddY = 51.035 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 0.95/2192.18 = 0 mm
Actual eccenY = My / P = 51.03/2192.18 = 23 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(0,22) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(23,20) = 23 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(23 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 16 nos. (5027 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00310 354.76 8.92 345.84 217.30 175 38.03
2 628 0.00270 350.18 8.92 341.26 214.42 125 26.80
3 628 0.00230 338.60 8.92 329.68 207.14 75 15.54
4 628 0.00189 323.19 8.90 314.30 197.48 25 4.94
5 628 0.00149 293.71 8.35 285.36 179.30 -25 -4.48
6 628 0.00109 218.28 7.08 211.20 132.70 -75 -9.95
7 628 0.00069 137.99 5.09 132.90 83.50 -125 -10.44
8 628 0.00029 57.71 2.39 55.32 34.76 -175 -6.08
Total1266.60 54.35
xux = 436 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 436/450 = 0.349
Puc = 0.349 x 20.00 x 300 x 450
= 941.63 kN
Pux1 = Puc + Pus(Total)
= 941.63 + (1266.60)
= 2208.23 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 436/450
= 0.403
Muc = 941.63 x (0.5 x 450 - 0.403 x 450) = 41.10 kN-m
Mux1 = Muc + Mus(Total)
= 41.10 + (54.35)
= 95.45 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 2513 0.00285 352.56 8.92 343.64 863.67 100 86.37
2 2513 0.00069 138.15 5.10 133.05 334.39 -100 -33.44
Total1198.06 52.93
xuy = 314 mm               Puc = C1.fck.B.D
ku = 1.047
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.976
C1 = 0.446 x (1 - C3/6) = 0.373
Puc = 0.373 x 20.00 x 450 x 300
= 1008.29 kN
Puy1 = Puc + Pus(Total)
= 1008.29 + (1198.06)
= 2206.35 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.431
Muc = 1008.29 x (0.5 x 300 - 0.431 x 300) = 20.99 kN-m
Muy1 = Muc + Mus(Total)
= 20.99 + (52.93)
= 73.92 kN-m
Pu/Puz = 0.802
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.802, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((47.68/95.45)2.000) + ((51.03/73.92)2.000)
= 0.726 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 20 = 320 mm
               iii) 300 mm
               Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG43 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 16 nos. (5027 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG44 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 414.31 kN
MomentX,(Mx) = 18.35 kN-m
MomentY,(My) = 92.10 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C520.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 8.576 kN.m
My_MinEccen = 8.286 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 18.351 kN.m
My = max(My,My_MinEccen) + MuaddY = 92.098 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 18.35/414.31 = 44 mm
Actual eccenY = My / P = 92.10/414.31 = 222 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(44,21) = 44 mm
eccenY = max(Actual eccenY,eccenYMin) = max(222,20) = 222 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(44 mm) > 0.05 x 300(15 mm)
and eccenY(222 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 2 nos. + #16 - 8 nos. (1835 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00271 350.52 8.92 341.60 137.36 177 24.31
2 402 0.00124 247.47 7.62 239.85 96.45 88 8.44
3 226 -0.00020 -40.50 0.00 -40.50 -9.16 0 0.00
4 402 -0.00164 -307.47 0.00 -307.47 -123.64 -88 10.82
5 402 -0.00312 -354.91 0.00 -354.91 -142.72 -177 25.26
Total-41.70 68.83
xux = 213 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 213/450 = 0.170
Puc = 0.170 x 20.00 x 300 x 450
= 459.42 kN
Pux1 = Puc + Pus(Total)
= 459.42 + (-41.70)
= 417.72 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 213/450
= 0.197
Muc = 459.42 x (0.5 x 450 - 0.197 x 450) = 62.72 kN-m
Mux1 = Muc + Mus(Total)
= 62.72 + (68.83)
= 131.55 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 917 0.00226 337.44 8.92 328.52 301.37 102 30.74
2 917 -0.00299 -353.80 0.00 -353.80 -324.55 -102 33.10
Total-23.19 63.84
xuy = 136 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 136/300 = 0.163
Puc = 0.163 x 20.00 x 450 x 300
= 440.44 kN
Puy1 = Puc + Pus(Total)
= 440.44 + (-23.19)
= 417.25 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 136/300
= 0.189
Muc = 440.44 x (0.5 x 300 - 0.189 x 300) = 41.16 kN-m
Muy1 = Muc + Mus(Total)
= 41.16 + (63.84)
= 105.00 kN-m
Pu/Puz = 0.234
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.234, an = 1.057
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((18.35/131.55)1.057) + ((92.10/105.00)1.057)
= 0.995 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG44 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 2 nos. + #16 - 8 nos. (1835 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 848.40 kN
MomentX,(Mx) = 17.35 kN-m
MomentY,(My) = 83.68 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C520.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 17.562 kN.m
My_MinEccen = 16.968 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 17.562 kN.m
My = max(My,My_MinEccen) + MuaddY = 83.679 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 17.35/848.40 = 20 mm
Actual eccenY = My / P = 83.68/848.40 = 99 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(20,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(99,20) = 99 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(99 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 6 nos. + #16 - 6 nos. (1885 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00294 353.41 8.92 344.49 138.53 177 24.52
2 226 0.00212 332.15 8.92 323.23 73.11 106 7.74
3 226 0.00132 263.81 7.89 255.92 57.89 37 2.12
4 402 0.00049 98.96 3.87 95.09 38.24 -35 -1.32
5 226 -0.00033 -65.88 0.00 -65.88 -14.90 -106 1.58
6 402 -0.00115 -230.73 0.00 -230.73 -92.78 -177 16.42
Total200.09 51.05
xux = 302 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 302/450 = 0.242
Puc = 0.242 x 20.00 x 300 x 450
= 653.06 kN
Pux1 = Puc + Pus(Total)
= 653.06 + (200.09)
= 853.15 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 302/450
= 0.280
Muc = 653.06 x (0.5 x 450 - 0.280 x 450) = 64.80 kN-m
Mux1 = Muc + Mus(Total)
= 64.80 + (51.05)
= 115.85 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 942 0.00269 350.00 8.92 341.08 321.46 102 32.79
2 942 -0.00075 -150.44 0.00 -150.44 -141.79 -102 14.46
Total179.67 47.25
xuy = 207 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 207/300 = 0.249
Puc = 0.249 x 20.00 x 450 x 300
= 672.05 kN
Puy1 = Puc + Pus(Total)
= 672.05 + (179.67)
= 851.72 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 207/300
= 0.288
Muc = 672.05 x (0.5 x 300 - 0.288 x 300) = 42.82 kN-m
Muy1 = Muc + Mus(Total)
= 42.82 + (47.25)
= 90.07 kN-m
Pu/Puz = 0.475
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.475, an = 1.459
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((17.56/115.85)1.459) + ((83.68/90.07)1.459)
= 0.962 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG44 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 6 nos. + #16 - 6 nos. (1885 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1279.08 kN
MomentX,(Mx) = 14.81 kN-m
MomentY,(My) = 80.80 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C520.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 26.477 kN.m
My_MinEccen = 25.582 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 26.477 kN.m
My = max(My,My_MinEccen) + MuaddY = 80.804 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 14.81/1279.08 = 12 mm
Actual eccenY = My / P = 80.80/1279.08 = 63 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(12,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(63,20) = 63 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(63 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 16 nos. (3217 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00300 353.94 8.92 345.02 138.74 177 24.56
2 402 0.00248 344.68 8.92 335.76 135.02 126 17.07
3 402 0.00196 326.32 8.92 317.40 127.64 76 9.68
4 402 0.00144 287.92 8.22 279.70 112.48 25 2.84
5 402 0.00092 183.58 6.31 177.27 71.28 -25 -1.80
6 402 0.00040 79.23 3.18 76.05 30.58 -76 -2.32
7 402 -0.00013 -25.11 0.00 -25.11 -10.10 -126 1.28
8 402 -0.00065 -129.46 0.00 -129.46 -52.06 -177 9.21
Total553.58 60.52
xux = 339 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 339/450 = 0.271
Puc = 0.271 x 20.00 x 300 x 450
= 732.80 kN
Pux1 = Puc + Pus(Total)
= 732.80 + (553.58)
= 1286.38 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 339/450
= 0.314
Muc = 732.80 x (0.5 x 450 - 0.314 x 450) = 61.46 kN-m
Mux1 = Muc + Mus(Total)
= 61.46 + (60.52)
= 121.98 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1608 0.00280 352.19 8.92 343.27 552.14 102 56.32
2 1608 -0.00015 -30.72 0.00 -30.72 -49.41 -102 5.04
Total502.73 61.36
xuy = 241 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 241/300 = 0.290
Puc = 0.290 x 20.00 x 450 x 300
= 782.16 kN
Puy1 = Puc + Pus(Total)
= 782.16 + (502.73)
= 1284.89 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 241/300
= 0.335
Muc = 782.16 x (0.5 x 300 - 0.335 x 300) = 38.78 kN-m
Muy1 = Muc + Mus(Total)
= 38.78 + (61.36)
= 100.13 kN-m
Pu/Puz = 0.585
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.585, an = 1.641
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((26.48/121.98)1.641) + ((80.80/100.13)1.641)
= 0.785 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG44 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 16 nos. (3217 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1706.07 kN
MomentX,(Mx) = 10.73 kN-m
MomentY,(My) = 76.60 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C520.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 35.316 kN.m
My_MinEccen = 34.121 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 35.316 kN.m
My = max(My,My_MinEccen) + MuaddY = 76.595 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 10.73/1706.07 = 6 mm
Actual eccenY = My / P = 76.60/1706.07 = 45 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(6,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(45,20) = 45 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(45 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 14 nos. (4398 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00303 354.19 8.92 345.27 216.94 175 37.96
2 628 0.00249 344.79 8.92 335.87 211.03 117 24.62
3 628 0.00194 325.61 8.91 316.70 198.99 58 11.61
4 628 0.00140 279.34 8.11 271.23 170.42 0 0.00
5 628 0.00085 170.28 5.98 164.31 103.24 -58 -6.02
6 628 0.00031 61.22 2.52 58.70 36.88 -117 -4.30
7 628 -0.00024 -47.84 0.00 -47.84 -30.06 -175 5.26
Total907.44 69.13
xux = 374 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 374/450 = 0.300
Puc = 0.300 x 20.00 x 300 x 450
= 808.73 kN
Pux1 = Puc + Pus(Total)
= 808.73 + (907.44)
= 1716.17 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 374/450
= 0.346
Muc = 808.73 x (0.5 x 450 - 0.346 x 450) = 56.00 kN-m
Mux1 = Muc + Mus(Total)
= 56.00 + (69.13)
= 125.13 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 2199 0.00285 352.54 8.92 343.62 755.67 100 75.57
2 2199 0.00023 45.03 1.90 43.13 94.86 -100 -9.49
Total850.53 66.08
xuy = 267 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 267/300 = 0.321
Puc = 0.321 x 20.00 x 450 x 300
= 865.69 kN
Puy1 = Puc + Pus(Total)
= 865.69 + (850.53)
= 1716.21 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 267/300
= 0.370
Muc = 865.69 x (0.5 x 300 - 0.370 x 300) = 33.63 kN-m
Muy1 = Muc + Mus(Total)
= 33.63 + (66.08)
= 99.71 kN-m
Pu/Puz = 0.671
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.671, an = 1.784
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((35.32/125.13)1.784) + ((76.60/99.71)1.784)
= 0.729 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 20 = 320 mm
               iii) 300 mm
               Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG44 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 14 nos. (4398 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 2127.99 kN
MomentX,(Mx) = 5.05 kN-m
MomentY,(My) = 49.86 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C520.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 46.284 kN.m
My_MinEccen = 42.560 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 46.284 kN.m
My = max(My,My_MinEccen) + MuaddY = 49.865 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 5.05/2127.99 = 2 mm
Actual eccenY = My / P = 49.86/2127.99 = 23 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(2,22) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(23,20) = 23 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(23 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 16 nos. (5027 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00309 354.65 8.92 345.73 217.23 175 38.01
2 628 0.00267 349.50 8.92 340.58 213.99 125 26.75
3 628 0.00226 337.13 8.92 328.21 206.22 75 15.47
4 628 0.00184 319.85 8.86 310.99 195.40 25 4.89
5 628 0.00143 285.19 8.19 277.00 174.04 -25 -4.35
6 628 0.00101 202.22 6.74 195.48 122.83 -75 -9.21
7 628 0.00060 119.26 4.53 114.73 72.09 -125 -9.01
8 628 0.00018 36.30 1.55 34.75 21.83 -175 -3.82
Total1223.63 58.72
xux = 422 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 422/450 = 0.338
Puc = 0.338 x 20.00 x 300 x 450
= 911.25 kN
Pux1 = Puc + Pus(Total)
= 911.25 + (1223.63)
= 2134.88 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 422/450
= 0.390
Muc = 911.25 x (0.5 x 450 - 0.390 x 450) = 45.11 kN-m
Mux1 = Muc + Mus(Total)
= 45.11 + (58.72)
= 103.83 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 2513 0.00290 353.06 8.92 344.14 864.92 100 86.49
2 2513 0.00060 120.49 4.56 115.92 291.35 -100 -29.13
Total1156.27 57.36
xuy = 302 mm               Puc = C1.fck.B.D
ku = 1.008
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.112
C1 = 0.446 x (1 - C3/6) = 0.363
Puc = 0.363 x 20.00 x 450 x 300
= 980.97 kN
Puy1 = Puc + Pus(Total)
= 980.97 + (1156.27)
= 2137.25 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.419
Muc = 980.97 x (0.5 x 300 - 0.419 x 300) = 23.92 kN-m
Muy1 = Muc + Mus(Total)
= 23.92 + (57.36)
= 81.27 kN-m
Pu/Puz = 0.778
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.778, an = 1.964
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((46.28/103.83)1.964) + ((49.86/81.27)1.964)
= 0.588 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 20 = 320 mm
               iii) 300 mm
               Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG44 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 16 nos. (5027 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG45 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 542.97 kN
MomentX,(Mx) = 16.00 kN-m
MomentY,(My) = 55.66 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C530.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 11.240 kN.m
My_MinEccen = 10.859 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 15.996 kN.m
My = max(My,My_MinEccen) + MuaddY = 55.663 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 16.00/542.97 = 29 mm
Actual eccenY = My / P = 55.66/542.97 = 103 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(29,21) = 29 mm
eccenY = max(Actual eccenY,eccenYMin) = max(103,20) = 103 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(29 mm) > 0.05 x 300(15 mm)
and eccenY(103 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00283 352.43 8.92 343.51 77.70 179 13.91
2 226 0.00153 297.29 8.43 288.86 65.34 90 5.85
3 226 0.00023 45.99 1.93 44.05 9.96 0 0.00
4 226 -0.00107 -214.17 0.00 -214.17 -48.44 -90 4.34
5 226 -0.00237 -341.39 0.00 -341.39 -77.22 -179 13.82
Total27.34 37.91
xux = 241 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 241/450 = 0.193
Puc = 0.193 x 20.00 x 300 x 450
= 520.17 kN
Pux1 = Puc + Pus(Total)
= 520.17 + (27.34)
= 547.51 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 241/450
= 0.223
Muc = 520.17 x (0.5 x 450 - 0.223 x 450) = 64.93 kN-m
Mux1 = Muc + Mus(Total)
= 64.93 + (37.91)
= 102.84 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00253 345.77 8.92 336.85 190.49 104 19.81
2 565 -0.00188 -322.32 0.00 -322.32 -182.27 -104 18.96
Total8.22 38.77
xuy = 165 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 165/300 = 0.198
Puc = 0.198 x 20.00 x 450 x 300
= 535.36 kN
Puy1 = Puc + Pus(Total)
= 535.36 + (8.22)
= 543.58 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 165/300
= 0.229
Muc = 535.36 x (0.5 x 300 - 0.229 x 300) = 43.50 kN-m
Muy1 = Muc + Mus(Total)
= 43.50 + (38.77)
= 82.27 kN-m
Pu/Puz = 0.349
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.349, an = 1.248
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((16.00/102.84)1.248) + ((55.66/82.27)1.248)
= 0.712 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG45 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1064.10 kN
MomentX,(Mx) = 13.80 kN-m
MomentY,(My) = 52.03 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C530.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 22.027 kN.m
My_MinEccen = 21.282 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 22.027 kN.m
My = max(My,My_MinEccen) + MuaddY = 52.029 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 13.80/1064.10 = 13 mm
Actual eccenY = My / P = 52.03/1064.10 = 49 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(13,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(49,20) = 49 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(49 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 4 nos. + #16 - 4 nos. (1257 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00306 354.44 8.92 345.52 138.94 177 24.59
2 226 0.00198 326.92 8.92 318.00 71.93 58 4.20
3 226 0.00091 182.43 6.28 176.15 39.84 -58 -2.32
4 402 -0.00017 -34.34 0.00 -34.34 -13.81 -177 2.44
Total236.91 28.91
xux = 383 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 383/450 = 0.307
Puc = 0.307 x 20.00 x 300 x 450
= 827.72 kN
Pux1 = Puc + Pus(Total)
= 827.72 + (236.91)
= 1064.63 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 383/450
= 0.354
Muc = 827.72 x (0.5 x 450 - 0.354 x 450) = 54.29 kN-m
Mux1 = Muc + Mus(Total)
= 54.29 + (28.91)
= 83.20 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00285 352.62 8.92 343.70 215.96 102 22.03
2 628 0.00011 21.95 0.95 20.99 13.19 -102 -1.35
Total229.15 20.68
xuy = 260 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 260/300 = 0.312
Puc = 0.312 x 20.00 x 450 x 300
= 842.91 kN
Puy1 = Puc + Pus(Total)
= 842.91 + (229.15)
= 1072.05 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 260/300
= 0.361
Muc = 842.91 x (0.5 x 300 - 0.361 x 300) = 35.21 kN-m
Muy1 = Muc + Mus(Total)
= 35.21 + (20.68)
= 55.89 kN-m
Pu/Puz = 0.667
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.667, an = 1.779
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((22.03/83.20)1.779) + ((52.03/55.89)1.779)
= 0.974 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG45 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 4 nos. + #16 - 4 nos. (1257 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1616.47 kN
MomentX,(Mx) = 10.28 kN-m
MomentY,(My) = 49.45 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C530.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 33.461 kN.m
My_MinEccen = 32.329 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 33.461 kN.m
My = max(My,My_MinEccen) + MuaddY = 49.449 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 10.28/1616.47 = 6 mm
Actual eccenY = My / P = 49.45/1616.47 = 31 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(6,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(31,20) = 31 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(31 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 14 nos. (2815 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00311 354.88 8.92 345.96 139.12 177 24.62
2 402 0.00263 348.56 8.92 339.64 136.58 118 16.12
3 402 0.00216 333.48 8.92 324.56 130.52 59 7.70
4 402 0.00168 309.75 8.69 301.06 121.06 0 0.00
5 402 0.00120 240.26 7.50 232.77 93.60 -59 -5.52
6 402 0.00072 144.76 5.29 139.47 56.08 -118 -6.62
7 402 0.00025 49.25 2.06 47.19 18.97 -177 -3.36
Total695.93 32.94
xux = 432 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 432/450 = 0.346
Puc = 0.346 x 20.00 x 300 x 450
= 934.03 kN
Pux1 = Puc + Pus(Total)
= 934.03 + (695.93)
= 1629.96 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 432/450
= 0.400
Muc = 934.03 x (0.5 x 450 - 0.400 x 450) = 42.14 kN-m
Mux1 = Muc + Mus(Total)
= 42.14 + (32.94)
= 75.08 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1407 0.00293 353.26 8.92 344.34 484.64 102 49.43
2 1407 0.00058 115.88 4.42 111.46 156.88 -102 -16.00
Total641.52 33.43
xuy = 302 mm               Puc = C1.fck.B.D
ku = 1.008
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.112
C1 = 0.446 x (1 - C3/6) = 0.363
Puc = 0.363 x 20.00 x 450 x 300
= 980.97 kN
Puy1 = Puc + Pus(Total)
= 980.97 + (641.52)
= 1622.49 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.419
Muc = 980.97 x (0.5 x 300 - 0.419 x 300) = 23.92 kN-m
Muy1 = Muc + Mus(Total)
= 23.92 + (33.43)
= 57.35 kN-m
Pu/Puz = 0.782
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.782, an = 1.971
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((33.46/75.08)1.971) + ((49.45/57.35)1.971)
= 0.950 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG45 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 14 nos. (2815 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 2221.23 kN
MomentX,(Mx) = 5.98 kN-m
MomentY,(My) = 44.92 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C530.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 45.979 kN.m
My_MinEccen = 44.425 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 45.979 kN.m
My = max(My,My_MinEccen) + MuaddY = 44.918 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 5.98/2221.23 = 3 mm
Actual eccenY = My / P = 44.92/2221.23 = 20 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(3,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(20,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 16 nos. (5027 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00310 354.82 8.92 345.90 217.33 175 38.03
2 628 0.00271 350.51 8.92 341.59 214.63 125 26.83
3 628 0.00231 339.30 8.92 330.38 207.59 75 15.57
4 628 0.00192 324.78 8.91 315.88 198.47 25 4.96
5 628 0.00152 296.72 8.42 288.31 181.15 -25 -4.53
6 628 0.00113 225.93 7.23 218.70 137.41 -75 -10.31
7 628 0.00073 146.91 5.35 141.56 88.95 -125 -11.12
8 628 0.00034 67.90 2.77 65.13 40.92 -175 -7.16
Total1286.45 52.28
xux = 443 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 443/450 = 0.354
Puc = 0.354 x 20.00 x 300 x 450
= 956.81 kN
Pux1 = Puc + Pus(Total)
= 956.81 + (1286.45)
= 2243.26 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 443/450
= 0.410
Muc = 956.81 x (0.5 x 450 - 0.410 x 450) = 38.97 kN-m
Mux1 = Muc + Mus(Total)
= 38.97 + (52.28)
= 91.24 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 2513 0.00283 352.38 8.92 343.46 863.21 100 86.32
2 2513 0.00072 144.60 5.28 139.32 350.14 -100 -35.01
Total1213.35 51.31
xuy = 319 mm               Puc = C1.fck.B.D
ku = 1.063
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.929
C1 = 0.446 x (1 - C3/6) = 0.377
Puc = 0.377 x 20.00 x 450 x 300
= 1017.83 kN
Puy1 = Puc + Pus(Total)
= 1017.83 + (1213.35)
= 2231.18 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.435
Muc = 1017.83 x (0.5 x 300 - 0.435 x 300) = 19.97 kN-m
Muy1 = Muc + Mus(Total)
= 19.97 + (51.31)
= 71.28 kN-m
Pu/Puz = 0.812
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.812, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((45.98/91.24)2.000) + ((44.92/71.28)2.000)
= 0.651 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 20 = 320 mm
               iii) 300 mm
               Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG45 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 16 nos. (5027 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 2933.45 kN
MomentX,(Mx) = 1.77 kN-m
MomentY,(My) = 27.22 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C530.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 63.803 kN.m
My_MinEccen = 58.669 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 63.803 kN.m
My = max(My,My_MinEccen) + MuaddY = 58.669 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 1.77/2933.45 = 1 mm
Actual eccenY = My / P = 27.22/2933.45 = 9 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(1,22) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(9,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #32 - 10 nos. (8042 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1608 0.00305 354.34 8.92 345.42 555.61 169 93.90
2 1608 0.00237 341.40 8.92 332.48 534.80 85 45.19
3 1608 0.00169 310.67 8.71 301.96 485.69 0 0.00
4 1608 0.00102 203.03 6.76 196.27 315.70 -85 -26.68
5 1608 0.00034 67.34 2.75 64.59 103.89 -169 -17.56
Total1995.69 94.85
xux = 436 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 436/450 = 0.349
Puc = 0.349 x 20.00 x 300 x 450
= 941.63 kN
Pux1 = Puc + Pus(Total)
= 941.63 + (1995.69)
= 2937.31 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 436/450
= 0.403
Muc = 941.63 x (0.5 x 450 - 0.403 x 450) = 41.10 kN-m
Mux1 = Muc + Mus(Total)
= 41.10 + (94.85)
= 135.96 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 4021 0.00280 352.17 8.92 343.25 1380.31 94 129.75
2 4021 0.00072 144.63 5.28 139.35 560.35 -94 -52.67
Total1940.66 77.08
xuy = 309 mm               Puc = C1.fck.B.D
ku = 1.031
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.027
C1 = 0.446 x (1 - C3/6) = 0.370
Puc = 0.370 x 20.00 x 450 x 300
= 998.00 kN
Puy1 = Puc + Pus(Total)
= 998.00 + (1940.66)
= 2938.66 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.426
Muc = 998.00 x (0.5 x 300 - 0.426 x 300) = 22.09 kN-m
Muy1 = Muc + Mus(Total)
= 22.09 + (77.08)
= 99.17 kN-m
Pu/Puz = 0.805
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.805, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((63.80/135.96)2.000) + ((58.67/99.17)2.000)
= 0.570 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 32/4 = 8 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (8, 6) = 8 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 32 = 512 mm
               iii) 300 mm
               Required spacing = minimum of (300, 512, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG45 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #32 - 10 nos. (8042 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG46 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 426.01 kN
MomentX,(Mx) = 5.87 kN-m
MomentY,(My) = 18.45 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C540.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 8.818 kN.m
My_MinEccen = 8.520 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 8.818 kN.m
My = max(My,My_MinEccen) + MuaddY = 18.455 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 5.87/426.01 = 14 mm
Actual eccenY = My / P = 18.45/426.01 = 43 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(14,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(43,20) = 43 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(43 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00274 351.20 8.92 342.28 77.42 179 13.86
2 226 0.00125 250.34 7.67 242.67 54.89 90 4.91
3 226 -0.00023 -46.67 0.00 -46.67 -10.56 0 0.00
4 226 -0.00172 -312.22 0.00 -312.22 -70.62 -90 6.32
5 226 -0.00320 -355.68 0.00 -355.68 -80.45 -179 14.40
Total-29.32 39.49
xux = 211 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 211/450 = 0.169
Puc = 0.169 x 20.00 x 300 x 450
= 455.63 kN
Pux1 = Puc + Pus(Total)
= 455.63 + (-29.32)
= 426.31 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 211/450
= 0.195
Muc = 455.63 x (0.5 x 450 - 0.195 x 450) = 62.53 kN-m
Mux1 = Muc + Mus(Total)
= 62.53 + (39.49)
= 102.03 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00232 339.33 8.92 330.41 186.84 104 19.43
2 565 -0.00304 -354.25 0.00 -354.25 -200.32 -104 20.83
Total-13.48 40.27
xuy = 136 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 136/300 = 0.163
Puc = 0.163 x 20.00 x 450 x 300
= 440.44 kN
Puy1 = Puc + Pus(Total)
= 440.44 + (-13.48)
= 426.96 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 136/300
= 0.189
Muc = 440.44 x (0.5 x 300 - 0.189 x 300) = 41.16 kN-m
Muy1 = Muc + Mus(Total)
= 41.16 + (40.27)
= 81.42 kN-m
Pu/Puz = 0.274
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.274, an = 1.123
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((8.82/102.03)1.123) + ((18.45/81.42)1.123)
= 0.253 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG46 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 885.47 kN
MomentX,(Mx) = 7.15 kN-m
MomentY,(My) = 16.77 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C540.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 18.329 kN.m
My_MinEccen = 17.709 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 18.329 kN.m
My = max(My,My_MinEccen) + MuaddY = 17.709 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 7.15/885.47 = 8 mm
Actual eccenY = My / P = 16.77/885.47 = 19 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(8,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(19,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00301 354.01 8.92 345.09 78.06 179 13.97
2 226 0.00206 330.12 8.92 321.20 72.65 90 6.50
3 226 0.00112 223.40 7.18 216.22 48.91 0 0.00
4 226 0.00017 33.83 1.44 32.38 7.32 -90 -0.66
5 226 -0.00078 -155.75 0.00 -155.75 -35.23 -179 6.31
Total171.71 26.13
xux = 330 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 330/450 = 0.264
Puc = 0.264 x 20.00 x 300 x 450
= 713.81 kN
Pux1 = Puc + Pus(Total)
= 713.81 + (171.71)
= 885.53 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 330/450
= 0.306
Muc = 713.81 x (0.5 x 450 - 0.306 x 450) = 62.48 kN-m
Mux1 = Muc + Mus(Total)
= 62.48 + (26.13)
= 88.60 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00280 352.11 8.92 343.19 194.07 104 20.18
2 565 -0.00039 -78.06 0.00 -78.06 -44.14 -104 4.59
Total149.92 24.77
xuy = 229 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 229/300 = 0.274
Puc = 0.274 x 20.00 x 450 x 300
= 740.39 kN
Puy1 = Puc + Pus(Total)
= 740.39 + (149.92)
= 890.32 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 229/300
= 0.317
Muc = 740.39 x (0.5 x 300 - 0.317 x 300) = 40.68 kN-m
Muy1 = Muc + Mus(Total)
= 40.68 + (24.77)
= 65.45 kN-m
Pu/Puz = 0.569
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.569, an = 1.615
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((18.33/88.60)1.615) + ((17.71/65.45)1.615)
= 0.200 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG46 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1320.25 kN
MomentX,(Mx) = 4.62 kN-m
MomentY,(My) = 13.90 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C540.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 27.329 kN.m
My_MinEccen = 26.405 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 27.329 kN.m
My = max(My,My_MinEccen) + MuaddY = 26.405 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 4.62/1320.25 = 3 mm
Actual eccenY = My / P = 13.90/1320.25 = 11 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(3,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(11,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 12 nos. (1357 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00313 355.01 8.92 346.09 78.28 179 14.01
2 226 0.00258 347.11 8.92 338.19 76.50 107 8.22
3 226 0.00203 328.77 8.92 319.85 72.35 36 2.59
4 226 0.00148 292.36 8.31 284.05 64.25 -36 -2.30
5 226 0.00093 185.86 6.36 179.50 40.60 -107 -4.36
6 226 0.00038 75.99 3.07 72.92 16.49 -179 -2.95
Total348.47 15.21
xux = 454 mm               Puc = C1.fck.B.D
ku = 1.008
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.112
C1 = 0.446 x (1 - C3/6) = 0.363
Puc = 0.363 x 20.00 x 300 x 450
= 980.97 kN
Pux1 = Puc + Pus(Total)
= 980.97 + (348.47)
= 1329.45 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.419
Muc = 980.97 x (0.5 x 450 - 0.419 x 450) = 35.88 kN-m
Mux1 = Muc + Mus(Total)
= 35.88 + (15.21)
= 51.08 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 679 0.00289 352.94 8.92 344.02 233.45 104 24.28
2 679 0.00065 129.52 4.84 124.68 84.61 -104 -8.80
Total318.05 15.48
xuy = 314 mm               Puc = C1.fck.B.D
ku = 1.047
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.976
C1 = 0.446 x (1 - C3/6) = 0.373
Puc = 0.373 x 20.00 x 450 x 300
= 1008.29 kN
Puy1 = Puc + Pus(Total)
= 1008.29 + (318.05)
= 1326.34 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.431
Muc = 1008.29 x (0.5 x 300 - 0.431 x 300) = 20.99 kN-m
Muy1 = Muc + Mus(Total)
= 20.99 + (15.48)
= 36.47 kN-m
Pu/Puz = 0.812
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.812, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((27.33/51.08)2.000) + ((26.41/36.47)2.000)
= 0.810 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG46 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 12 nos. (1357 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1712.58 kN
MomentX,(Mx) = 1.61 kN-m
MomentY,(My) = 9.13 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C540.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 35.450 kN.m
My_MinEccen = 34.252 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 35.450 kN.m
My = max(My,My_MinEccen) + MuaddY = 34.252 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 1.61/1712.58 = 1 mm
Actual eccenY = My / P = 9.13/1712.58 = 5 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(1,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(5,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 14 nos. (2815 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00310 354.75 8.92 345.83 139.06 177 24.61
2 402 0.00265 348.97 8.92 340.05 136.74 118 16.14
3 402 0.00220 335.21 8.92 326.29 131.21 59 7.74
4 402 0.00176 314.60 8.79 305.82 122.98 0 0.00
5 402 0.00131 262.00 7.86 254.14 102.19 -59 -6.03
6 402 0.00086 172.66 6.04 166.62 67.00 -118 -7.91
7 402 0.00042 83.33 3.33 80.00 32.17 -177 -5.69
Total731.36 28.86
xux = 457 mm               Puc = C1.fck.B.D
ku = 1.016
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.083
C1 = 0.446 x (1 - C3/6) = 0.366
Puc = 0.366 x 20.00 x 300 x 450
= 986.88 kN
Pux1 = Puc + Pus(Total)
= 986.88 + (731.36)
= 1718.23 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.421
Muc = 986.88 x (0.5 x 450 - 0.421 x 450) = 34.93 kN-m
Mux1 = Muc + Mus(Total)
= 34.93 + (28.86)
= 63.79 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1407 0.00281 352.22 8.92 343.30 483.17 102 49.28
2 1407 0.00076 152.59 5.51 147.08 207.01 -102 -21.11
Total690.18 28.17
xuy = 328 mm               Puc = C1.fck.B.D
ku = 1.094
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.843
C1 = 0.446 x (1 - C3/6) = 0.383
Puc = 0.383 x 20.00 x 450 x 300
= 1034.93 kN
Puy1 = Puc + Pus(Total)
= 1034.93 + (690.18)
= 1725.10 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.442
Muc = 1034.93 x (0.5 x 300 - 0.442 x 300) = 18.14 kN-m
Muy1 = Muc + Mus(Total)
= 18.14 + (28.17)
= 46.30 kN-m
Pu/Puz = 0.829
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.829, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((35.45/63.79)2.000) + ((34.25/46.30)2.000)
= 0.856 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG46 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 14 nos. (2815 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 2014.72 kN
MomentX,(Mx) = 1.73 kN-m
MomentY,(My) = 3.19 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C540.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 43.820 kN.m
My_MinEccen = 40.294 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 43.820 kN.m
My = max(My,My_MinEccen) + MuaddY = 40.294 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 1.73/2014.72 = 1 mm
Actual eccenY = My / P = 3.19/2014.72 = 2 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(1,22) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(2,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 14 nos. (4398 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00310 354.73 8.92 345.81 217.28 175 38.02
2 628 0.00262 348.28 8.92 339.36 213.23 117 24.88
3 628 0.00215 333.29 8.92 324.37 203.81 58 11.89
4 628 0.00168 309.75 8.69 301.06 189.16 0 0.00
5 628 0.00121 241.34 7.52 233.83 146.92 -58 -8.57
6 628 0.00073 146.91 5.35 141.56 88.95 -117 -10.38
7 628 0.00026 52.48 2.19 50.30 31.60 -175 -5.53
Total1090.94 50.31
xux = 432 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 432/450 = 0.346
Puc = 0.346 x 20.00 x 300 x 450
= 934.03 kN
Pux1 = Puc + Pus(Total)
= 934.03 + (1090.94)
= 2024.97 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 432/450
= 0.400
Muc = 934.03 x (0.5 x 450 - 0.400 x 450) = 42.14 kN-m
Mux1 = Muc + Mus(Total)
= 42.14 + (50.31)
= 92.45 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 2199 0.00287 352.75 8.92 343.83 756.13 100 75.61
2 2199 0.00066 131.36 4.90 126.46 278.10 -100 -27.81
Total1034.24 47.80
xuy = 309 mm               Puc = C1.fck.B.D
ku = 1.031
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.027
C1 = 0.446 x (1 - C3/6) = 0.370
Puc = 0.370 x 20.00 x 450 x 300
= 998.00 kN
Puy1 = Puc + Pus(Total)
= 998.00 + (1034.24)
= 2032.23 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.426
Muc = 998.00 x (0.5 x 300 - 0.426 x 300) = 22.09 kN-m
Muy1 = Muc + Mus(Total)
= 22.09 + (47.80)
= 69.90 kN-m
Pu/Puz = 0.792
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.792, an = 1.986
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((43.82/92.45)1.986) + ((40.29/69.90)1.986)
= 0.562 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 20 = 320 mm
               iii) 300 mm
               Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG46 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 14 nos. (4398 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG47 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 379.41 kN
MomentX,(Mx) = 3.74 kN-m
MomentY,(My) = 0.31 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C550.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 7.854 kN.m
My_MinEccen = 7.588 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 7.854 kN.m
My = max(My,My_MinEccen) + MuaddY = 7.588 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 3.74/379.41 = 10 mm
Actual eccenY = My / P = 0.31/379.41 = 1 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(10,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(1,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00269 350.08 8.92 341.16 77.17 179 13.81
2 226 0.00112 224.59 7.20 217.38 49.17 90 4.40
3 226 -0.00045 -89.43 0.00 -89.43 -20.23 0 0.00
4 226 -0.00202 -328.37 0.00 -328.37 -74.28 -90 6.65
5 226 -0.00359 -359.04 0.00 -359.04 -81.21 -179 14.54
Total-49.38 39.40
xux = 200 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 200/450 = 0.160
Puc = 0.160 x 20.00 x 300 x 450
= 430.95 kN
Pux1 = Puc + Pus(Total)
= 430.95 + (-49.38)
= 381.57 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 200/450
= 0.184
Muc = 430.95 x (0.5 x 450 - 0.184 x 450) = 61.20 kN-m
Mux1 = Muc + Mus(Total)
= 61.20 + (39.40)
= 100.59 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00220 335.00 8.92 326.08 184.40 104 19.18
2 565 -0.00369 -359.94 0.00 -359.94 -203.54 -104 21.17
Total-19.15 40.35
xuy = 124 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 124/300 = 0.148
Puc = 0.148 x 20.00 x 450 x 300
= 400.57 kN
Puy1 = Puc + Pus(Total)
= 400.57 + (-19.15)
= 381.42 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 124/300
= 0.171
Muc = 400.57 x (0.5 x 300 - 0.171 x 300) = 39.48 kN-m
Muy1 = Muc + Mus(Total)
= 39.48 + (40.35)
= 79.83 kN-m
Pu/Puz = 0.244
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.244, an = 1.073
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((7.85/100.59)1.073) + ((7.59/79.83)1.073)
= 0.145 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG47 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 775.38 kN
MomentX,(Mx) = 0.27 kN-m
MomentY,(My) = 0.62 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C550.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 16.050 kN.m
My_MinEccen = 15.508 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 16.050 kN.m
My = max(My,My_MinEccen) + MuaddY = 15.508 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 0.27/775.38 = 0 mm
Actual eccenY = My / P = 0.62/775.38 = 1 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(0,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(1,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00297 353.62 8.92 344.70 77.97 179 13.96
2 226 0.00193 325.22 8.91 316.31 71.55 90 6.40
3 226 0.00090 179.07 6.20 172.87 39.10 0 0.00
4 226 -0.00014 -28.14 0.00 -28.14 -6.37 -90 0.57
5 226 -0.00118 -235.36 0.00 -235.36 -53.24 -179 9.53
Total129.02 30.46
xux = 302 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 302/450 = 0.242
Puc = 0.242 x 20.00 x 300 x 450
= 653.06 kN
Pux1 = Puc + Pus(Total)
= 653.06 + (129.02)
= 782.08 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 302/450
= 0.280
Muc = 653.06 x (0.5 x 450 - 0.280 x 450) = 64.80 kN-m
Mux1 = Muc + Mus(Total)
= 64.80 + (30.46)
= 95.26 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00273 350.98 8.92 342.06 193.43 104 20.12
2 565 -0.00076 -152.37 0.00 -152.37 -86.17 -104 8.96
Total107.27 29.08
xuy = 209 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 209/300 = 0.250
Puc = 0.250 x 20.00 x 450 x 300
= 675.84 kN
Puy1 = Puc + Pus(Total)
= 675.84 + (107.27)
= 783.11 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 209/300
= 0.289
Muc = 675.84 x (0.5 x 300 - 0.289 x 300) = 42.73 kN-m
Muy1 = Muc + Mus(Total)
= 42.73 + (29.08)
= 71.81 kN-m
Pu/Puz = 0.498
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.498, an = 1.497
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((16.05/95.26)1.497) + ((15.51/71.81)1.497)
= 0.170 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG47 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1181.79 kN
MomentX,(Mx) = 1.85 kN-m
MomentY,(My) = 1.04 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C550.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 24.463 kN.m
My_MinEccen = 23.636 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 24.463 kN.m
My = max(My,My_MinEccen) + MuaddY = 23.636 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 1.85/1181.79 = 2 mm
Actual eccenY = My / P = 1.04/1181.79 = 1 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(2,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(1,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00312 354.96 8.92 346.04 78.27 179 14.01
2 226 0.00239 341.89 8.92 332.97 75.32 90 6.74
3 226 0.00165 307.87 8.64 299.23 67.68 0 0.00
4 226 0.00091 182.48 6.28 176.19 39.85 -90 -3.57
5 226 0.00018 35.20 1.50 33.70 7.62 -179 -1.36
Total268.75 15.82
xux = 425 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 425/450 = 0.340
Puc = 0.340 x 20.00 x 300 x 450
= 918.84 kN
Pux1 = Puc + Pus(Total)
= 918.84 + (268.75)
= 1187.59 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 425/450
= 0.393
Muc = 918.84 x (0.5 x 450 - 0.393 x 450) = 44.14 kN-m
Mux1 = Muc + Mus(Total)
= 44.14 + (15.82)
= 59.96 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00295 353.43 8.92 344.51 194.81 104 20.26
2 565 0.00044 88.22 3.50 84.71 47.90 -104 -4.98
Total242.72 15.28
xuy = 291 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 291/300 = 0.349
Puc = 0.349 x 20.00 x 450 x 300
= 941.63 kN
Puy1 = Puc + Pus(Total)
= 941.63 + (242.72)
= 1184.34 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 291/300
= 0.403
Muc = 941.63 x (0.5 x 300 - 0.403 x 300) = 27.40 kN-m
Muy1 = Muc + Mus(Total)
= 27.40 + (15.28)
= 42.68 kN-m
Pu/Puz = 0.759
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.759, an = 1.932
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((24.46/59.96)1.932) + ((23.64/42.68)1.932)
= 0.496 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG47 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1611.23 kN
MomentX,(Mx) = 2.43 kN-m
MomentY,(My) = 1.51 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C550.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 33.352 kN.m
My_MinEccen = 32.225 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 33.352 kN.m
My = max(My,My_MinEccen) + MuaddY = 32.225 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 2.43/1611.23 = 2 mm
Actual eccenY = My / P = 1.51/1611.23 = 1 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(2,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(1,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 12 nos. (2413 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00308 354.62 8.92 345.70 139.01 177 24.61
2 402 0.00255 346.48 8.92 337.56 135.74 106 14.42
3 402 0.00202 328.63 8.92 319.71 128.56 35 4.55
4 402 0.00150 293.95 8.35 285.59 114.84 -35 -4.07
5 402 0.00097 193.28 6.54 186.74 75.09 -106 -7.97
6 402 0.00044 87.48 3.48 84.01 33.78 -177 -5.98
Total627.04 25.55
xux = 461 mm               Puc = C1.fck.B.D
ku = 1.023
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.055
C1 = 0.446 x (1 - C3/6) = 0.368
Puc = 0.368 x 20.00 x 300 x 450
= 992.55 kN
Pux1 = Puc + Pus(Total)
= 992.55 + (627.04)
= 1619.59 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.424
Muc = 992.55 x (0.5 x 450 - 0.424 x 450) = 34.02 kN-m
Mux1 = Muc + Mus(Total)
= 34.02 + (25.55)
= 59.57 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1206 0.00281 352.22 8.92 343.30 414.14 102 42.24
2 1206 0.00076 152.59 5.51 147.08 177.44 -102 -18.10
Total591.58 24.14
xuy = 328 mm               Puc = C1.fck.B.D
ku = 1.094
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.843
C1 = 0.446 x (1 - C3/6) = 0.383
Puc = 0.383 x 20.00 x 450 x 300
= 1034.93 kN
Puy1 = Puc + Pus(Total)
= 1034.93 + (591.58)
= 1626.51 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.442
Muc = 1034.93 x (0.5 x 300 - 0.442 x 300) = 18.14 kN-m
Muy1 = Muc + Mus(Total)
= 18.14 + (24.14)
= 42.28 kN-m
Pu/Puz = 0.829
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.829, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((33.35/59.57)2.000) + ((32.22/42.28)2.000)
= 0.894 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG47 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 12 nos. (2413 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 2098.95 kN
MomentX,(Mx) = 3.48 kN-m
MomentY,(My) = 1.37 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C550.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 45.652 kN.m
My_MinEccen = 41.979 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 45.652 kN.m
My = max(My,My_MinEccen) + MuaddY = 41.979 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 3.48/2098.95 = 2 mm
Actual eccenY = My / P = 1.37/2098.95 = 1 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(2,22) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(1,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 14 nos. (4398 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00311 354.87 8.92 345.95 217.37 175 38.04
2 628 0.00266 349.16 8.92 340.24 213.78 117 24.94
3 628 0.00220 335.22 8.92 326.30 205.02 58 11.96
4 628 0.00175 314.19 8.78 305.41 191.89 0 0.00
5 628 0.00130 259.26 7.82 251.44 157.99 -58 -9.22
6 628 0.00084 168.52 5.93 162.59 102.16 -117 -11.92
7 628 0.00039 77.78 3.13 74.65 46.90 -175 -8.21
Total1135.11 45.60
xux = 450 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 450/450 = 0.360
Puc = 0.360 x 20.00 x 300 x 450
= 972.00 kN
Pux1 = Puc + Pus(Total)
= 972.00 + (1135.11)
= 2107.11 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 450/450
= 0.416
Muc = 972.00 x (0.5 x 450 - 0.416 x 450) = 36.74 kN-m
Mux1 = Muc + Mus(Total)
= 36.74 + (45.60)
= 82.34 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 2199 0.00281 352.21 8.92 343.29 754.93 100 75.49
2 2199 0.00075 150.74 5.46 145.29 319.51 -100 -31.95
Total1074.43 43.54
xuy = 323 mm               Puc = C1.fck.B.D
ku = 1.078
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.884
C1 = 0.446 x (1 - C3/6) = 0.380
Puc = 0.380 x 20.00 x 450 x 300
= 1026.69 kN
Puy1 = Puc + Pus(Total)
= 1026.69 + (1074.43)
= 2101.12 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.438
Muc = 1026.69 x (0.5 x 300 - 0.438 x 300) = 19.02 kN-m
Muy1 = Muc + Mus(Total)
= 19.02 + (43.54)
= 62.56 kN-m
Pu/Puz = 0.825
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.825, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((45.65/82.34)2.000) + ((41.98/62.56)2.000)
= 0.758 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 20 = 320 mm
               iii) 300 mm
               Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG47 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 14 nos. (4398 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG48 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 424.93 kN
MomentX,(Mx) = 9.84 kN-m
MomentY,(My) = 15.31 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C560.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 8.796 kN.m
My_MinEccen = 8.499 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 9.837 kN.m
My = max(My,My_MinEccen) + MuaddY = 15.312 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 9.84/424.93 = 23 mm
Actual eccenY = My / P = 15.31/424.93 = 36 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(23,21) = 23 mm
eccenY = max(Actual eccenY,eccenYMin) = max(36,20) = 36 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(23 mm) > 0.05 x 300(15 mm)
and eccenY(36 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00274 351.20 8.92 342.28 77.42 179 13.86
2 226 0.00125 250.34 7.67 242.67 54.89 90 4.91
3 226 -0.00023 -46.67 0.00 -46.67 -10.56 0 0.00
4 226 -0.00172 -312.22 0.00 -312.22 -70.62 -90 6.32
5 226 -0.00320 -355.68 0.00 -355.68 -80.45 -179 14.40
Total-29.32 39.49
xux = 211 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 211/450 = 0.169
Puc = 0.169 x 20.00 x 300 x 450
= 455.63 kN
Pux1 = Puc + Pus(Total)
= 455.63 + (-29.32)
= 426.31 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 211/450
= 0.195
Muc = 455.63 x (0.5 x 450 - 0.195 x 450) = 62.53 kN-m
Mux1 = Muc + Mus(Total)
= 62.53 + (39.49)
= 102.03 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00232 339.33 8.92 330.41 186.84 104 19.43
2 565 -0.00304 -354.25 0.00 -354.25 -200.32 -104 20.83
Total-13.48 40.27
xuy = 136 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 136/300 = 0.163
Puc = 0.163 x 20.00 x 450 x 300
= 440.44 kN
Puy1 = Puc + Pus(Total)
= 440.44 + (-13.48)
= 426.96 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 136/300
= 0.189
Muc = 440.44 x (0.5 x 300 - 0.189 x 300) = 41.16 kN-m
Muy1 = Muc + Mus(Total)
= 41.16 + (40.27)
= 81.42 kN-m
Pu/Puz = 0.273
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.273, an = 1.122
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((9.84/102.03)1.122) + ((15.31/81.42)1.122)
= 0.226 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG48 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 881.73 kN
MomentX,(Mx) = 6.57 kN-m
MomentY,(My) = 13.20 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C560.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 18.252 kN.m
My_MinEccen = 17.635 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 18.252 kN.m
My = max(My,My_MinEccen) + MuaddY = 17.635 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 6.57/881.73 = 7 mm
Actual eccenY = My / P = 13.20/881.73 = 15 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(7,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(15,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00301 354.01 8.92 345.09 78.06 179 13.97
2 226 0.00206 330.12 8.92 321.20 72.65 90 6.50
3 226 0.00112 223.40 7.18 216.22 48.91 0 0.00
4 226 0.00017 33.83 1.44 32.38 7.32 -90 -0.66
5 226 -0.00078 -155.75 0.00 -155.75 -35.23 -179 6.31
Total171.71 26.13
xux = 330 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 330/450 = 0.264
Puc = 0.264 x 20.00 x 300 x 450
= 713.81 kN
Pux1 = Puc + Pus(Total)
= 713.81 + (171.71)
= 885.53 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 330/450
= 0.306
Muc = 713.81 x (0.5 x 450 - 0.306 x 450) = 62.48 kN-m
Mux1 = Muc + Mus(Total)
= 62.48 + (26.13)
= 88.60 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00279 352.08 8.92 343.16 194.05 104 20.18
2 565 -0.00041 -82.08 0.00 -82.08 -46.41 -104 4.83
Total147.64 25.01
xuy = 227 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 227/300 = 0.273
Puc = 0.273 x 20.00 x 450 x 300
= 736.59 kN
Puy1 = Puc + Pus(Total)
= 736.59 + (147.64)
= 884.23 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 227/300
= 0.315
Muc = 736.59 x (0.5 x 300 - 0.315 x 300) = 40.83 kN-m
Muy1 = Muc + Mus(Total)
= 40.83 + (25.01)
= 65.83 kN-m
Pu/Puz = 0.566
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.566, an = 1.611
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((18.25/88.60)1.611) + ((17.63/65.83)1.611)
= 0.198 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG48 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1314.46 kN
MomentX,(Mx) = 3.62 kN-m
MomentY,(My) = 9.87 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C560.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 27.209 kN.m
My_MinEccen = 26.289 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 27.209 kN.m
My = max(My,My_MinEccen) + MuaddY = 26.289 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 3.62/1314.46 = 3 mm
Actual eccenY = My / P = 9.87/1314.46 = 8 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(3,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(8,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 4 nos. + #16 - 4 nos. (1257 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00304 354.26 8.92 345.34 138.87 177 24.58
2 226 0.00219 334.65 8.92 325.73 73.68 58 4.30
3 226 0.00135 269.93 7.98 261.95 59.25 -58 -3.46
4 402 0.00050 99.33 3.88 95.45 38.38 -177 -6.79
Total310.18 18.63
xux = 471 mm               Puc = C1.fck.B.D
ku = 1.047
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.976
C1 = 0.446 x (1 - C3/6) = 0.373
Puc = 0.373 x 20.00 x 300 x 450
= 1008.29 kN
Pux1 = Puc + Pus(Total)
= 1008.29 + (310.18)
= 1318.47 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.431
Muc = 1008.29 x (0.5 x 450 - 0.431 x 450) = 31.49 kN-m
Mux1 = Muc + Mus(Total)
= 31.49 + (18.63)
= 50.11 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00285 352.56 8.92 343.64 215.92 102 22.02
2 628 0.00070 140.39 5.16 135.23 84.97 -102 -8.67
Total300.89 13.36
xuy = 319 mm               Puc = C1.fck.B.D
ku = 1.063
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.929
C1 = 0.446 x (1 - C3/6) = 0.377
Puc = 0.377 x 20.00 x 450 x 300
= 1017.83 kN
Puy1 = Puc + Pus(Total)
= 1017.83 + (300.89)
= 1318.71 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.435
Muc = 1017.83 x (0.5 x 300 - 0.435 x 300) = 19.97 kN-m
Muy1 = Muc + Mus(Total)
= 19.97 + (13.36)
= 33.33 kN-m
Pu/Puz = 0.824
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.824, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((27.21/50.11)2.000) + ((26.29/33.33)2.000)
= 0.917 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG48 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 4 nos. + #16 - 4 nos. (1257 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1704.94 kN
MomentX,(Mx) = 2.19 kN-m
MomentY,(My) = 4.84 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C560.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 35.292 kN.m
My_MinEccen = 34.099 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 35.292 kN.m
My = max(My,My_MinEccen) + MuaddY = 34.099 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 2.19/1704.94 = 1 mm
Actual eccenY = My / P = 4.84/1704.94 = 3 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(1,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(3,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 14 nos. (2815 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00310 354.75 8.92 345.83 139.06 177 24.61
2 402 0.00265 348.97 8.92 340.05 136.74 118 16.14
3 402 0.00220 335.21 8.92 326.29 131.21 59 7.74
4 402 0.00176 314.60 8.79 305.82 122.98 0 0.00
5 402 0.00131 262.00 7.86 254.14 102.19 -59 -6.03
6 402 0.00086 172.66 6.04 166.62 67.00 -118 -7.91
7 402 0.00042 83.33 3.33 80.00 32.17 -177 -5.69
Total731.36 28.86
xux = 457 mm               Puc = C1.fck.B.D
ku = 1.016
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.083
C1 = 0.446 x (1 - C3/6) = 0.366
Puc = 0.366 x 20.00 x 300 x 450
= 986.88 kN
Pux1 = Puc + Pus(Total)
= 986.88 + (731.36)
= 1718.23 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.421
Muc = 986.88 x (0.5 x 450 - 0.421 x 450) = 34.93 kN-m
Mux1 = Muc + Mus(Total)
= 34.93 + (28.86)
= 63.79 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1407 0.00283 352.39 8.92 343.47 483.41 102 49.31
2 1407 0.00073 146.64 5.34 141.30 198.87 -102 -20.28
Total682.27 29.02
xuy = 323 mm               Puc = C1.fck.B.D
ku = 1.078
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.884
C1 = 0.446 x (1 - C3/6) = 0.380
Puc = 0.380 x 20.00 x 450 x 300
= 1026.69 kN
Puy1 = Puc + Pus(Total)
= 1026.69 + (682.27)
= 1708.96 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.438
Muc = 1026.69 x (0.5 x 300 - 0.438 x 300) = 19.02 kN-m
Muy1 = Muc + Mus(Total)
= 19.02 + (29.02)
= 48.04 kN-m
Pu/Puz = 0.825
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.825, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((35.29/63.79)2.000) + ((34.10/48.04)2.000)
= 0.810 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG48 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 14 nos. (2815 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 2004.99 kN
MomentX,(Mx) = 0.61 kN-m
MomentY,(My) = 0.71 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C560.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 43.609 kN.m
My_MinEccen = 40.100 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 43.609 kN.m
My = max(My,My_MinEccen) + MuaddY = 40.100 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 0.61/2004.99 = 0 mm
Actual eccenY = My / P = 0.71/2004.99 = 0 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(0,22) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(0,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 14 nos. (4398 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00309 354.70 8.92 345.78 217.26 175 38.02
2 628 0.00262 348.10 8.92 339.18 213.11 117 24.86
3 628 0.00214 332.88 8.92 323.96 203.55 58 11.87
4 628 0.00166 308.82 8.67 300.15 188.59 0 0.00
5 628 0.00119 237.58 7.45 230.13 144.60 -58 -8.43
6 628 0.00071 142.38 5.22 137.16 86.18 -117 -10.05
7 628 0.00024 47.18 1.98 45.20 28.40 -175 -4.97
Total1081.69 51.30
xux = 429 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 429/450 = 0.343
Puc = 0.343 x 20.00 x 300 x 450
= 926.44 kN
Pux1 = Puc + Pus(Total)
= 926.44 + (1081.69)
= 2008.13 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 429/450
= 0.397
Muc = 926.44 x (0.5 x 450 - 0.397 x 450) = 43.15 kN-m
Mux1 = Muc + Mus(Total)
= 43.15 + (51.30)
= 94.45 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 2199 0.00289 352.96 8.92 344.04 756.58 100 75.66
2 2199 0.00062 124.21 4.68 119.53 262.86 -100 -26.29
Total1019.43 49.37
xuy = 305 mm               Puc = C1.fck.B.D
ku = 1.016
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.083
C1 = 0.446 x (1 - C3/6) = 0.366
Puc = 0.366 x 20.00 x 450 x 300
= 986.88 kN
Puy1 = Puc + Pus(Total)
= 986.88 + (1019.43)
= 2006.31 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.421
Muc = 986.88 x (0.5 x 300 - 0.421 x 300) = 23.28 kN-m
Muy1 = Muc + Mus(Total)
= 23.28 + (49.37)
= 72.66 kN-m
Pu/Puz = 0.788
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.788, an = 1.980
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((43.61/94.45)1.980) + ((40.10/72.66)1.980)
= 0.525 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 20 = 320 mm
               iii) 300 mm
               Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG48 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 14 nos. (4398 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG49 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 530.35 kN
MomentX,(Mx) = 17.91 kN-m
MomentY,(My) = 49.77 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C570.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 10.978 kN.m
My_MinEccen = 10.607 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 17.909 kN.m
My = max(My,My_MinEccen) + MuaddY = 49.766 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 17.91/530.35 = 34 mm
Actual eccenY = My / P = 49.77/530.35 = 94 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(34,21) = 34 mm
eccenY = max(Actual eccenY,eccenYMin) = max(94,20) = 94 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(34 mm) > 0.05 x 300(15 mm)
and eccenY(94 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00282 352.34 8.92 343.42 77.68 179 13.90
2 226 0.00150 294.53 8.37 286.16 64.73 90 5.79
3 226 0.00018 36.30 1.55 34.75 7.86 0 0.00
4 226 -0.00114 -227.71 0.00 -227.71 -51.51 -90 4.61
5 226 -0.00246 -344.05 0.00 -344.05 -77.82 -179 13.93
Total20.94 38.24
xux = 237 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 237/450 = 0.190
Puc = 0.190 x 20.00 x 300 x 450
= 512.58 kN
Pux1 = Puc + Pus(Total)
= 512.58 + (20.94)
= 533.52 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 237/450
= 0.219
Muc = 512.58 x (0.5 x 450 - 0.219 x 450) = 64.73 kN-m
Mux1 = Muc + Mus(Total)
= 64.73 + (38.24)
= 102.97 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00251 345.41 8.92 336.49 190.28 104 19.79
2 565 -0.00196 -326.18 0.00 -326.18 -184.45 -104 19.18
Total5.83 38.97
xuy = 163 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 163/300 = 0.195
Puc = 0.195 x 20.00 x 450 x 300
= 527.77 kN
Puy1 = Puc + Pus(Total)
= 527.77 + (5.83)
= 533.60 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 163/300
= 0.226
Muc = 527.77 x (0.5 x 300 - 0.226 x 300) = 43.40 kN-m
Muy1 = Muc + Mus(Total)
= 43.40 + (38.97)
= 82.37 kN-m
Pu/Puz = 0.341
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.341, an = 1.234
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((17.91/102.97)1.234) + ((49.77/82.37)1.234)
= 0.652 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG49 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1043.34 kN
MomentX,(Mx) = 14.43 kN-m
MomentY,(My) = 46.04 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C570.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 21.597 kN.m
My_MinEccen = 20.867 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 21.597 kN.m
My = max(My,My_MinEccen) + MuaddY = 46.045 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 14.43/1043.34 = 14 mm
Actual eccenY = My / P = 46.04/1043.34 = 44 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(14,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(44,20) = 44 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(44 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00308 354.56 8.92 345.64 78.18 179 13.99
2 226 0.00225 336.96 8.92 328.04 74.20 90 6.64
3 226 0.00143 285.19 8.19 277.00 62.66 0 0.00
4 226 0.00060 120.18 4.55 115.63 26.15 -90 -2.34
5 226 -0.00022 -44.82 0.00 -44.82 -10.14 -179 1.81
Total231.05 20.11
xux = 380 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 380/450 = 0.304
Puc = 0.304 x 20.00 x 300 x 450
= 820.13 kN
Pux1 = Puc + Pus(Total)
= 820.13 + (231.05)
= 1051.18 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 380/450
= 0.351
Muc = 820.13 x (0.5 x 450 - 0.351 x 450) = 54.99 kN-m
Mux1 = Muc + Mus(Total)
= 54.99 + (20.11)
= 75.10 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00288 352.86 8.92 343.94 194.49 104 20.23
2 565 0.00008 16.56 0.72 15.84 8.96 -104 -0.93
Total203.45 19.30
xuy = 260 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 260/300 = 0.312
Puc = 0.312 x 20.00 x 450 x 300
= 842.91 kN
Puy1 = Puc + Pus(Total)
= 842.91 + (203.45)
= 1046.36 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 260/300
= 0.361
Muc = 842.91 x (0.5 x 300 - 0.361 x 300) = 35.21 kN-m
Muy1 = Muc + Mus(Total)
= 35.21 + (19.30)
= 54.51 kN-m
Pu/Puz = 0.670
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.670, an = 1.784
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((21.60/75.10)1.784) + ((46.04/54.51)1.784)
= 0.848 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG49 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1583.34 kN
MomentX,(Mx) = 10.73 kN-m
MomentY,(My) = 43.00 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C570.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 32.775 kN.m
My_MinEccen = 31.667 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 32.775 kN.m
My = max(My,My_MinEccen) + MuaddY = 43.004 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 10.73/1583.34 = 7 mm
Actual eccenY = My / P = 43.00/1583.34 = 27 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(7,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(27,20) = 27 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(27 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 14 nos. (2815 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00310 354.79 8.92 345.87 139.08 177 24.62
2 402 0.00261 348.00 8.92 339.08 136.35 118 16.09
3 402 0.00212 332.25 8.92 323.33 130.02 59 7.67
4 402 0.00163 306.91 8.62 298.29 119.95 0 0.00
5 402 0.00114 228.77 7.29 221.48 89.06 -59 -5.25
6 402 0.00065 130.87 4.88 125.99 50.66 -118 -5.98
7 402 0.00016 32.98 1.41 31.57 12.69 -177 -2.25
Total677.83 34.90
xux = 422 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 422/450 = 0.338
Puc = 0.338 x 20.00 x 300 x 450
= 911.25 kN
Pux1 = Puc + Pus(Total)
= 911.25 + (677.83)
= 1589.08 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 422/450
= 0.390
Muc = 911.25 x (0.5 x 450 - 0.390 x 450) = 45.11 kN-m
Mux1 = Muc + Mus(Total)
= 45.11 + (34.90)
= 80.01 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1407 0.00294 353.34 8.92 344.42 484.74 102 49.44
2 1407 0.00054 107.37 4.15 103.22 145.28 -102 -14.82
Total630.02 34.63
xuy = 298 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 298/300 = 0.357
Puc = 0.357 x 20.00 x 450 x 300
= 964.41 kN
Puy1 = Puc + Pus(Total)
= 964.41 + (630.02)
= 1594.43 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 298/300
= 0.413
Muc = 964.41 x (0.5 x 300 - 0.413 x 300) = 25.24 kN-m
Muy1 = Muc + Mus(Total)
= 25.24 + (34.63)
= 59.87 kN-m
Pu/Puz = 0.766
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.766, an = 1.944
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((32.78/80.01)1.944) + ((43.00/59.87)1.944)
= 0.702 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG49 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 14 nos. (2815 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 2169.40 kN
MomentX,(Mx) = 7.41 kN-m
MomentY,(My) = 38.20 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C570.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 44.907 kN.m
My_MinEccen = 43.388 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 44.907 kN.m
My = max(My,My_MinEccen) + MuaddY = 43.388 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 7.41/2169.40 = 3 mm
Actual eccenY = My / P = 38.20/2169.40 = 18 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(3,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(18,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 16 nos. (5027 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00309 354.70 8.92 345.78 217.26 175 38.02
2 628 0.00268 349.84 8.92 340.92 214.21 125 26.78
3 628 0.00228 337.88 8.92 328.96 206.69 75 15.50
4 628 0.00187 321.55 8.88 312.67 196.46 25 4.91
5 628 0.00146 290.59 8.27 282.32 177.39 -25 -4.43
6 628 0.00105 210.38 6.92 203.47 127.84 -75 -9.59
7 628 0.00064 128.78 4.82 123.96 77.89 -125 -9.74
8 628 0.00024 47.18 1.98 45.20 28.40 -175 -4.97
Total1246.13 56.48
xux = 429 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 429/450 = 0.343
Puc = 0.343 x 20.00 x 300 x 450
= 926.44 kN
Pux1 = Puc + Pus(Total)
= 926.44 + (1246.13)
= 2172.57 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 429/450
= 0.397
Muc = 926.44 x (0.5 x 450 - 0.397 x 450) = 43.15 kN-m
Mux1 = Muc + Mus(Total)
= 43.15 + (56.48)
= 99.63 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 2513 0.00287 352.75 8.92 343.83 864.15 100 86.42
2 2513 0.00066 131.36 4.90 126.46 317.83 -100 -31.78
Total1181.98 54.63
xuy = 309 mm               Puc = C1.fck.B.D
ku = 1.031
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.027
C1 = 0.446 x (1 - C3/6) = 0.370
Puc = 0.370 x 20.00 x 450 x 300
= 998.00 kN
Puy1 = Puc + Pus(Total)
= 998.00 + (1181.98)
= 2179.98 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.426
Muc = 998.00 x (0.5 x 300 - 0.426 x 300) = 22.09 kN-m
Muy1 = Muc + Mus(Total)
= 22.09 + (54.63)
= 76.72 kN-m
Pu/Puz = 0.793
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.793, an = 1.989
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((44.91/99.63)1.989) + ((43.39/76.72)1.989)
= 0.527 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 20 = 320 mm
               iii) 300 mm
               Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG49 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 16 nos. (5027 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 2850.96 kN
MomentX,(Mx) = 3.11 kN-m
MomentY,(My) = 22.42 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C570.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 62.008 kN.m
My_MinEccen = 57.019 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 62.008 kN.m
My = max(My,My_MinEccen) + MuaddY = 57.019 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 3.11/2850.96 = 1 mm
Actual eccenY = My / P = 22.42/2850.96 = 8 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(1,22) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(8,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #32 - 10 nos. (8042 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1608 0.00304 354.24 8.92 345.32 555.45 169 93.87
2 1608 0.00234 340.38 8.92 331.46 533.14 85 45.05
3 1608 0.00165 307.87 8.64 299.23 481.30 0 0.00
4 1608 0.00095 190.70 6.48 184.23 296.33 -85 -25.04
5 1608 0.00026 51.65 2.16 49.50 79.62 -169 -13.46
Total1945.85 100.43
xux = 425 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 425/450 = 0.340
Puc = 0.340 x 20.00 x 300 x 450
= 918.84 kN
Pux1 = Puc + Pus(Total)
= 918.84 + (1945.85)
= 2864.69 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 425/450
= 0.393
Muc = 918.84 x (0.5 x 450 - 0.393 x 450) = 44.14 kN-m
Mux1 = Muc + Mus(Total)
= 44.14 + (100.43)
= 144.57 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 4021 0.00285 352.56 8.92 343.64 1381.85 94 129.89
2 4021 0.00065 130.67 4.88 125.79 505.83 -94 -47.55
Total1887.69 82.35
xuy = 300 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 300/300 = 0.360
Puc = 0.360 x 20.00 x 450 x 300
= 972.00 kN
Puy1 = Puc + Pus(Total)
= 972.00 + (1887.69)
= 2859.69 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 300/300
= 0.416
Muc = 972.00 x (0.5 x 300 - 0.416 x 300) = 24.49 kN-m
Muy1 = Muc + Mus(Total)
= 24.49 + (82.35)
= 106.84 kN-m
Pu/Puz = 0.782
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.782, an = 1.970
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((62.01/144.57)1.970) + ((57.02/106.84)1.970)
= 0.479 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 32/4 = 8 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (8, 6) = 8 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 32 = 512 mm
               iii) 300 mm
               Required spacing = minimum of (300, 512, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG49 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #32 - 10 nos. (8042 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG50 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 395.85 kN
MomentX,(Mx) = 14.42 kN-m
MomentY,(My) = 86.94 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C580.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 8.194 kN.m
My_MinEccen = 7.917 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 14.421 kN.m
My = max(My,My_MinEccen) + MuaddY = 86.936 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 14.42/395.85 = 36 mm
Actual eccenY = My / P = 86.94/395.85 = 220 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(36,21) = 36 mm
eccenY = max(Actual eccenY,eccenYMin) = max(220,20) = 220 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(36 mm) > 0.05 x 300(15 mm)
and eccenY(220 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 4 nos. + #16 - 6 nos. (1659 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00269 350.00 8.92 341.08 137.16 177 24.28
2 226 0.00120 239.34 7.48 231.86 52.45 89 4.64
3 402 -0.00030 -59.32 0.00 -59.32 -23.85 0 0.00
4 226 -0.00179 -316.68 0.00 -316.68 -71.63 -89 6.34
5 402 -0.00328 -356.38 0.00 -356.38 -143.31 -177 25.37
Total-49.19 60.62
xux = 207 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 207/450 = 0.166
Puc = 0.166 x 20.00 x 300 x 450
= 448.03 kN
Pux1 = Puc + Pus(Total)
= 448.03 + (-49.19)
= 398.84 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 207/450
= 0.192
Muc = 448.03 x (0.5 x 450 - 0.192 x 450) = 62.15 kN-m
Mux1 = Muc + Mus(Total)
= 62.15 + (60.62)
= 122.77 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 829 0.00221 335.40 8.92 326.48 270.77 102 27.62
2 829 -0.00328 -356.35 0.00 -356.35 -295.55 -102 30.15
Total-24.78 57.77
xuy = 130 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 130/300 = 0.156
Puc = 0.156 x 20.00 x 450 x 300
= 421.45 kN
Puy1 = Puc + Pus(Total)
= 421.45 + (-24.78)
= 396.67 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 130/300
= 0.180
Muc = 421.45 x (0.5 x 300 - 0.180 x 300) = 40.41 kN-m
Muy1 = Muc + Mus(Total)
= 40.41 + (57.77)
= 98.18 kN-m
Pu/Puz = 0.231
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.231, an = 1.051
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((14.42/122.77)1.051) + ((86.94/98.18)1.051)
= 0.985 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG50 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 4 nos. + #16 - 6 nos. (1659 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 811.20 kN
MomentX,(Mx) = 14.13 kN-m
MomentY,(My) = 79.85 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C580.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 16.792 kN.m
My_MinEccen = 16.224 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 16.792 kN.m
My = max(My,My_MinEccen) + MuaddY = 79.853 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 14.13/811.20 = 17 mm
Actual eccenY = My / P = 79.85/811.20 = 98 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(17,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(98,20) = 98 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(98 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 4 nos. + #16 - 6 nos. (1659 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00294 353.36 8.92 344.44 138.51 177 24.52
2 226 0.00190 323.63 8.90 314.73 71.19 89 6.30
3 402 0.00086 172.94 6.05 166.90 67.11 0 0.00
4 226 -0.00017 -34.37 0.00 -34.37 -7.77 -89 0.69
5 402 -0.00121 -241.68 0.00 -241.68 -97.18 -177 17.20
Total171.85 48.71
xux = 299 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 299/450 = 0.239
Puc = 0.239 x 20.00 x 300 x 450
= 645.47 kN
Pux1 = Puc + Pus(Total)
= 645.47 + (171.85)
= 817.32 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 299/450
= 0.276
Muc = 645.47 x (0.5 x 450 - 0.276 x 450) = 64.99 kN-m
Mux1 = Muc + Mus(Total)
= 64.99 + (48.71)
= 113.70 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 829 0.00268 349.76 8.92 340.84 282.69 102 28.83
2 829 -0.00080 -160.16 0.00 -160.16 -132.83 -102 13.55
Total149.86 42.38
xuy = 205 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 205/300 = 0.246
Puc = 0.246 x 20.00 x 450 x 300
= 664.45 kN
Puy1 = Puc + Pus(Total)
= 664.45 + (149.86)
= 814.31 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 205/300
= 0.284
Muc = 664.45 x (0.5 x 300 - 0.284 x 300) = 42.98 kN-m
Muy1 = Muc + Mus(Total)
= 42.98 + (42.38)
= 85.37 kN-m
Pu/Puz = 0.473
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.473, an = 1.454
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((16.79/113.70)1.454) + ((79.85/85.37)1.454)
= 0.969 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG50 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 4 nos. + #16 - 6 nos. (1659 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1225.75 kN
MomentX,(Mx) = 11.75 kN-m
MomentY,(My) = 77.50 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C580.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 25.373 kN.m
My_MinEccen = 24.515 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 25.373 kN.m
My = max(My,My_MinEccen) + MuaddY = 77.495 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 11.75/1225.75 = 10 mm
Actual eccenY = My / P = 77.50/1225.75 = 63 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(10,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(63,20) = 63 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(63 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 14 nos. (2815 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00301 353.99 8.92 345.07 138.76 177 24.56
2 402 0.00241 342.71 8.92 333.79 134.22 118 15.84
3 402 0.00181 317.62 8.84 308.79 124.17 59 7.33
4 402 0.00120 240.51 7.50 233.01 93.70 0 0.00
5 402 0.00060 120.03 4.55 115.48 46.44 -59 -2.74
6 402 -0.00000 -0.46 0.00 -0.46 -0.19 -118 0.02
7 402 -0.00060 -120.95 0.00 -120.95 -48.64 -177 8.61
Total488.47 53.62
xux = 343 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 343/450 = 0.274
Puc = 0.274 x 20.00 x 300 x 450
= 740.39 kN
Pux1 = Puc + Pus(Total)
= 740.39 + (488.47)
= 1228.86 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 343/450
= 0.317
Muc = 740.39 x (0.5 x 450 - 0.317 x 450) = 61.01 kN-m
Mux1 = Muc + Mus(Total)
= 61.01 + (53.62)
= 114.63 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1407 0.00281 352.22 8.92 343.30 483.17 102 49.28
2 1407 -0.00014 -27.19 0.00 -27.19 -38.27 -102 3.90
Total444.90 53.19
xuy = 243 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 243/300 = 0.291
Puc = 0.291 x 20.00 x 450 x 300
= 785.95 kN
Puy1 = Puc + Pus(Total)
= 785.95 + (444.90)
= 1230.85 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 243/300
= 0.336
Muc = 785.95 x (0.5 x 300 - 0.336 x 300) = 38.58 kN-m
Muy1 = Muc + Mus(Total)
= 38.58 + (53.19)
= 91.77 kN-m
Pu/Puz = 0.593
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.593, an = 1.656
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((25.37/114.63)1.656) + ((77.50/91.77)1.656)
= 0.838 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG50 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 14 nos. (2815 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1640.21 kN
MomentX,(Mx) = 7.74 kN-m
MomentY,(My) = 73.79 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C580.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 33.952 kN.m
My_MinEccen = 32.804 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 33.952 kN.m
My = max(My,My_MinEccen) + MuaddY = 73.786 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 7.74/1640.21 = 5 mm
Actual eccenY = My / P = 73.79/1640.21 = 45 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(5,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(45,20) = 45 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(45 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 12 nos. (3770 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00305 354.32 8.92 345.40 217.02 175 37.98
2 628 0.00241 342.90 8.92 333.98 209.85 105 22.03
3 628 0.00178 316.09 8.81 307.27 193.07 35 6.76
4 628 0.00115 229.37 7.30 222.08 139.53 -35 -4.88
5 628 0.00051 102.67 3.99 98.68 62.00 -105 -6.51
6 628 -0.00012 -24.04 0.00 -24.04 -15.11 -175 2.64
Total806.36 58.02
xux = 387 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 387/450 = 0.309
Puc = 0.309 x 20.00 x 300 x 450
= 835.31 kN
Pux1 = Puc + Pus(Total)
= 835.31 + (806.36)
= 1641.67 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 387/450
= 0.358
Muc = 835.31 x (0.5 x 450 - 0.358 x 450) = 53.56 kN-m
Mux1 = Muc + Mus(Total)
= 53.56 + (58.02)
= 111.58 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1885 0.00286 352.69 8.92 343.77 647.99 100 64.80
2 1885 0.00031 61.82 2.54 59.28 111.74 -100 -11.17
Total759.73 53.63
xuy = 274 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 274/300 = 0.329
Puc = 0.329 x 20.00 x 450 x 300
= 888.47 kN
Puy1 = Puc + Pus(Total)
= 888.47 + (759.73)
= 1648.20 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 274/300
= 0.380
Muc = 888.47 x (0.5 x 300 - 0.380 x 300) = 31.92 kN-m
Muy1 = Muc + Mus(Total)
= 31.92 + (53.63)
= 85.54 kN-m
Pu/Puz = 0.697
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.697, an = 1.828
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((33.95/111.58)1.828) + ((73.79/85.54)1.828)
= 0.877 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 20 = 320 mm
               iii) 300 mm
               Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG50 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 12 nos. (3770 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 2054.89 kN
MomentX,(Mx) = 4.25 kN-m
MomentY,(My) = 48.35 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C580.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 44.694 kN.m
My_MinEccen = 41.098 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 44.694 kN.m
My = max(My,My_MinEccen) + MuaddY = 48.346 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 4.25/2054.89 = 2 mm
Actual eccenY = My / P = 48.35/2054.89 = 24 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(2,22) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(24,20) = 24 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(24 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 14 nos. (4398 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00310 354.82 8.92 345.90 217.33 175 38.03
2 628 0.00264 348.82 8.92 339.90 213.56 117 24.92
3 628 0.00218 334.47 8.92 325.55 204.55 58 11.93
4 628 0.00172 312.46 8.75 303.71 190.83 0 0.00
5 628 0.00126 252.26 7.70 244.56 153.66 -58 -8.96
6 628 0.00080 160.08 5.71 154.37 96.99 -117 -11.32
7 628 0.00034 67.90 2.77 65.13 40.92 -175 -7.16
Total1117.85 47.44
xux = 443 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 443/450 = 0.354
Puc = 0.354 x 20.00 x 300 x 450
= 956.81 kN
Pux1 = Puc + Pus(Total)
= 956.81 + (1117.85)
= 2074.66 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 443/450
= 0.410
Muc = 956.81 x (0.5 x 450 - 0.410 x 450) = 38.97 kN-m
Mux1 = Muc + Mus(Total)
= 38.97 + (47.44)
= 86.41 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 2199 0.00285 352.56 8.92 343.64 755.71 100 75.57
2 2199 0.00069 138.15 5.10 133.05 292.59 -100 -29.26
Total1048.30 46.31
xuy = 314 mm               Puc = C1.fck.B.D
ku = 1.047
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.976
C1 = 0.446 x (1 - C3/6) = 0.373
Puc = 0.373 x 20.00 x 450 x 300
= 1008.29 kN
Puy1 = Puc + Pus(Total)
= 1008.29 + (1048.30)
= 2056.59 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.431
Muc = 1008.29 x (0.5 x 300 - 0.431 x 300) = 20.99 kN-m
Muy1 = Muc + Mus(Total)
= 20.99 + (46.31)
= 67.30 kN-m
Pu/Puz = 0.808
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.808, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((44.69/86.41)2.000) + ((48.35/67.30)2.000)
= 0.784 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 20 = 320 mm
               iii) 300 mm
               Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG50 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 14 nos. (4398 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG51 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 406.80 kN
MomentX,(Mx) = 0.85 kN-m
MomentY,(My) = 82.83 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C590.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 8.421 kN.m
My_MinEccen = 8.136 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 8.421 kN.m
My = max(My,My_MinEccen) + MuaddY = 82.826 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 0.85/406.80 = 2 mm
Actual eccenY = My / P = 82.83/406.80 = 204 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(2,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(204,20) = 204 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(204 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 2 nos. + #16 - 6 nos. (1433 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00272 350.77 8.92 341.85 137.46 177 24.33
2 226 0.00081 162.54 5.78 156.76 35.46 60 2.12
3 402 -0.00109 -218.89 0.00 -218.89 -88.02 -58 5.08
4 402 -0.00303 -354.20 0.00 -354.20 -142.43 -177 25.21
Total-57.53 56.73
xux = 215 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 215/450 = 0.172
Puc = 0.172 x 20.00 x 300 x 450
= 465.12 kN
Pux1 = Puc + Pus(Total)
= 465.12 + (-57.53)
= 407.59 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 215/450
= 0.199
Muc = 465.12 x (0.5 x 450 - 0.199 x 450) = 62.99 kN-m
Mux1 = Muc + Mus(Total)
= 62.99 + (56.73)
= 119.72 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 716 0.00223 336.24 8.92 327.32 234.45 102 23.91
2 716 -0.00316 -355.30 0.00 -355.30 -254.50 -102 25.96
Total-20.05 49.87
xuy = 132 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 132/300 = 0.159
Puc = 0.159 x 20.00 x 450 x 300
= 429.05 kN
Puy1 = Puc + Pus(Total)
= 429.05 + (-20.05)
= 409.00 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 132/300
= 0.184
Muc = 429.05 x (0.5 x 300 - 0.184 x 300) = 40.72 kN-m
Muy1 = Muc + Mus(Total)
= 40.72 + (49.87)
= 90.59 kN-m
Pu/Puz = 0.247
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.247, an = 1.078
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((8.42/119.72)1.078) + ((82.83/90.59)1.078)
= 0.965 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG51 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 2 nos. + #16 - 6 nos. (1433 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 838.45 kN
MomentX,(Mx) = 2.13 kN-m
MomentY,(My) = 74.50 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C590.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 17.356 kN.m
My_MinEccen = 16.769 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 17.356 kN.m
My = max(My,My_MinEccen) + MuaddY = 74.495 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 2.13/838.45 = 3 mm
Actual eccenY = My / P = 74.50/838.45 = 89 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(3,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(89,20) = 89 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(89 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 14 nos. (1583 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00297 353.62 8.92 344.70 77.97 179 13.96
2 226 0.00228 337.91 8.92 328.99 74.41 119 8.88
3 226 0.00159 302.54 8.54 294.00 66.50 60 3.97
4 226 0.00090 179.07 6.20 172.87 39.10 -0 -0.00
5 226 0.00020 40.93 1.73 39.19 8.87 -60 -0.53
6 226 -0.00049 -97.22 0.00 -97.22 -21.99 -119 2.62
7 226 -0.00118 -235.36 0.00 -235.36 -53.24 -179 9.53
Total191.63 38.43
xux = 302 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 302/450 = 0.242
Puc = 0.242 x 20.00 x 300 x 450
= 653.06 kN
Pux1 = Puc + Pus(Total)
= 653.06 + (191.63)
= 844.69 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 302/450
= 0.280
Muc = 653.06 x (0.5 x 450 - 0.280 x 450) = 64.80 kN-m
Mux1 = Muc + Mus(Total)
= 64.80 + (38.43)
= 103.23 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 792 0.00274 351.20 8.92 342.28 270.98 104 28.18
2 792 -0.00071 -142.90 0.00 -142.90 -113.13 -104 11.77
Total157.84 39.95
xuy = 211 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 211/300 = 0.253
Puc = 0.253 x 20.00 x 450 x 300
= 683.44 kN
Puy1 = Puc + Pus(Total)
= 683.44 + (157.84)
= 841.28 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 211/300
= 0.293
Muc = 683.44 x (0.5 x 300 - 0.293 x 300) = 42.54 kN-m
Muy1 = Muc + Mus(Total)
= 42.54 + (39.95)
= 82.49 kN-m
Pu/Puz = 0.495
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.495, an = 1.492
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((17.36/103.23)1.492) + ((74.50/82.49)1.492)
= 0.929 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG51 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 14 nos. (1583 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1255.85 kN
MomentX,(Mx) = 2.12 kN-m
MomentY,(My) = 71.17 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C590.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 25.996 kN.m
My_MinEccen = 25.117 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 25.996 kN.m
My = max(My,My_MinEccen) + MuaddY = 71.170 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 2.12/1255.85 = 2 mm
Actual eccenY = My / P = 71.17/1255.85 = 57 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(2,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(57,20) = 57 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(57 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 14 nos. (2815 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00302 354.05 8.92 345.13 138.79 177 24.57
2 402 0.00242 343.16 8.92 334.24 134.41 118 15.86
3 402 0.00183 319.23 8.86 310.37 124.81 59 7.36
4 402 0.00124 247.47 7.62 239.85 96.45 0 0.00
5 402 0.00064 128.81 4.82 123.99 49.86 -59 -2.94
6 402 0.00005 10.15 0.45 9.70 3.90 -118 -0.46
7 402 -0.00054 -108.51 0.00 -108.51 -43.64 -177 7.72
Total504.58 52.11
xux = 348 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 348/450 = 0.278
Puc = 0.278 x 20.00 x 300 x 450
= 751.78 kN
Pux1 = Puc + Pus(Total)
= 751.78 + (504.58)
= 1256.36 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 348/450
= 0.322
Muc = 751.78 x (0.5 x 450 - 0.322 x 450) = 60.30 kN-m
Mux1 = Muc + Mus(Total)
= 60.30 + (52.11)
= 112.41 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1407 0.00282 352.30 8.92 343.38 483.29 102 49.30
2 1407 -0.00008 -16.80 0.00 -16.80 -23.64 -102 2.41
Total459.64 51.71
xuy = 246 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 246/300 = 0.295
Puc = 0.295 x 20.00 x 450 x 300
= 797.34 kN
Puy1 = Puc + Pus(Total)
= 797.34 + (459.64)
= 1256.99 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 246/300
= 0.341
Muc = 797.34 x (0.5 x 300 - 0.341 x 300) = 37.97 kN-m
Muy1 = Muc + Mus(Total)
= 37.97 + (51.71)
= 89.68 kN-m
Pu/Puz = 0.608
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.608, an = 1.680
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((26.00/112.41)1.680) + ((71.17/89.68)1.680)
= 0.764 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG51 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 14 nos. (2815 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1649.74 kN
MomentX,(Mx) = 2.63 kN-m
MomentY,(My) = 66.16 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C590.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 34.150 kN.m
My_MinEccen = 32.995 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 34.150 kN.m
My = max(My,My_MinEccen) + MuaddY = 66.160 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 2.63/1649.74 = 2 mm
Actual eccenY = My / P = 66.16/1649.74 = 40 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(2,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(40,20) = 40 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(40 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 18 nos. (3619 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00307 354.54 8.92 345.62 138.98 177 24.60
2 402 0.00268 349.74 8.92 340.82 137.05 133 18.19
3 402 0.00229 338.27 8.92 329.35 132.44 89 11.72
4 402 0.00189 323.14 8.89 314.24 126.36 44 5.59
5 402 0.00150 294.38 8.36 286.02 115.02 0 0.00
6 402 0.00111 221.33 7.14 214.19 86.13 -44 -3.81
7 402 0.00071 142.67 5.23 137.44 55.27 -89 -4.89
8 402 0.00032 64.00 2.63 61.37 24.68 -133 -3.28
9 402 -0.00007 -14.67 0.00 -14.67 -5.90 -177 1.04
Total810.04 49.17
xux = 394 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 394/450 = 0.315
Puc = 0.315 x 20.00 x 300 x 450
= 850.50 kN
Pux1 = Puc + Pus(Total)
= 850.50 + (810.04)
= 1660.54 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 394/450
= 0.364
Muc = 850.50 x (0.5 x 450 - 0.364 x 450) = 52.05 kN-m
Mux1 = Muc + Mus(Total)
= 52.05 + (49.17)
= 101.22 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1810 0.00290 353.05 8.92 344.13 622.72 102 63.52
2 1810 0.00036 72.80 2.95 69.85 126.40 -102 -12.89
Total749.12 50.63
xuy = 281 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 281/300 = 0.338
Puc = 0.338 x 20.00 x 450 x 300
= 911.25 kN
Puy1 = Puc + Pus(Total)
= 911.25 + (749.12)
= 1660.37 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 281/300
= 0.390
Muc = 911.25 x (0.5 x 300 - 0.390 x 300) = 30.07 kN-m
Muy1 = Muc + Mus(Total)
= 30.07 + (50.63)
= 80.70 kN-m
Pu/Puz = 0.715
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.715, an = 1.858
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((34.15/101.22)1.858) + ((66.16/80.70)1.858)
= 0.824 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG51 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 18 nos. (3619 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 1994.69 kN
MomentX,(Mx) = 2.34 kN-m
MomentY,(My) = 42.87 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C590.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 43.385 kN.m
My_MinEccen = 39.894 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 43.385 kN.m
My = max(My,My_MinEccen) + MuaddY = 42.873 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 2.34/1994.69 = 1 mm
Actual eccenY = My / P = 42.87/1994.69 = 21 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(1,22) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(21,20) = 21 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(21 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 14 nos. (4398 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00309 354.70 8.92 345.78 217.26 175 38.02
2 628 0.00262 348.10 8.92 339.18 213.11 117 24.86
3 628 0.00214 332.88 8.92 323.96 203.55 58 11.87
4 628 0.00166 308.82 8.67 300.15 188.59 0 0.00
5 628 0.00119 237.58 7.45 230.13 144.60 -58 -8.43
6 628 0.00071 142.38 5.22 137.16 86.18 -117 -10.05
7 628 0.00024 47.18 1.98 45.20 28.40 -175 -4.97
Total1081.69 51.30
xux = 429 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 429/450 = 0.343
Puc = 0.343 x 20.00 x 300 x 450
= 926.44 kN
Pux1 = Puc + Pus(Total)
= 926.44 + (1081.69)
= 2008.13 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 429/450
= 0.397
Muc = 926.44 x (0.5 x 450 - 0.397 x 450) = 43.15 kN-m
Mux1 = Muc + Mus(Total)
= 43.15 + (51.30)
= 94.45 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 2199 0.00289 352.96 8.92 344.04 756.58 100 75.66
2 2199 0.00062 124.21 4.68 119.53 262.86 -100 -26.29
Total1019.43 49.37
xuy = 305 mm               Puc = C1.fck.B.D
ku = 1.016
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.083
C1 = 0.446 x (1 - C3/6) = 0.366
Puc = 0.366 x 20.00 x 450 x 300
= 986.88 kN
Puy1 = Puc + Pus(Total)
= 986.88 + (1019.43)
= 2006.31 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.421
Muc = 986.88 x (0.5 x 300 - 0.421 x 300) = 23.28 kN-m
Muy1 = Muc + Mus(Total)
= 23.28 + (49.37)
= 72.66 kN-m
Pu/Puz = 0.784
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.784, an = 1.973
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((43.38/94.45)1.973) + ((42.87/72.66)1.973)
= 0.569 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 20 = 320 mm
               iii) 300 mm
               Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG51 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 14 nos. (4398 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG52 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 552.01 kN
MomentX,(Mx) = 5.94 kN-m
MomentY,(My) = 44.59 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C600.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 11.427 kN.m
My_MinEccen = 11.040 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 11.427 kN.m
My = max(My,My_MinEccen) + MuaddY = 44.594 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 5.94/552.01 = 11 mm
Actual eccenY = My / P = 44.59/552.01 = 81 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(11,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(81,20) = 81 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(81 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00284 352.47 8.92 343.55 77.71 179 13.91
2 226 0.00154 298.64 8.46 290.19 65.64 90 5.87
3 226 0.00025 50.72 2.12 48.61 10.99 0 0.00
4 226 -0.00104 -207.54 0.00 -207.54 -46.95 -90 4.20
5 226 -0.00233 -339.83 0.00 -339.83 -76.87 -179 13.76
Total30.53 37.75
xux = 243 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 243/450 = 0.194
Puc = 0.194 x 20.00 x 300 x 450
= 523.97 kN
Pux1 = Puc + Pus(Total)
= 523.97 + (30.53)
= 554.50 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 243/450
= 0.224
Muc = 523.97 x (0.5 x 450 - 0.224 x 450) = 65.02 kN-m
Mux1 = Muc + Mus(Total)
= 65.02 + (37.75)
= 102.76 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00254 346.12 8.92 337.20 190.68 104 19.83
2 565 -0.00180 -317.62 0.00 -317.62 -179.61 -104 18.68
Total11.07 38.51
xuy = 168 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 168/300 = 0.201
Puc = 0.201 x 20.00 x 450 x 300
= 542.95 kN
Puy1 = Puc + Pus(Total)
= 542.95 + (11.07)
= 554.03 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 168/300
= 0.232
Muc = 542.95 x (0.5 x 300 - 0.232 x 300) = 43.59 kN-m
Muy1 = Muc + Mus(Total)
= 43.59 + (38.51)
= 82.10 kN-m
Pu/Puz = 0.355
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.355, an = 1.258
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((11.43/102.76)1.258) + ((44.59/82.10)1.258)
= 0.527 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG52 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1098.85 kN
MomentX,(Mx) = 6.73 kN-m
MomentY,(My) = 41.28 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C600.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 22.746 kN.m
My_MinEccen = 21.977 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 22.746 kN.m
My = max(My,My_MinEccen) + MuaddY = 41.282 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 6.73/1098.85 = 6 mm
Actual eccenY = My / P = 41.28/1098.85 = 38 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(6,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(38,20) = 38 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(38 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00309 354.73 8.92 345.81 78.22 179 14.00
2 226 0.00231 338.99 8.92 330.07 74.66 90 6.68
3 226 0.00152 296.06 8.40 287.66 65.07 0 0.00
4 226 0.00073 145.84 5.32 140.52 31.78 -90 -2.84
5 226 -0.00006 -11.87 0.00 -11.87 -2.68 -179 0.48
Total247.05 18.32
xux = 397 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 397/450 = 0.318
Puc = 0.318 x 20.00 x 300 x 450
= 858.09 kN
Pux1 = Puc + Pus(Total)
= 858.09 + (247.05)
= 1105.14 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 397/450
= 0.367
Muc = 858.09 x (0.5 x 450 - 0.367 x 450) = 51.26 kN-m
Mux1 = Muc + Mus(Total)
= 51.26 + (18.32)
= 69.58 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00291 353.09 8.92 344.17 194.63 104 20.24
2 565 0.00023 46.02 1.93 44.09 24.93 -104 -2.59
Total219.56 17.65
xuy = 272 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 272/300 = 0.326
Puc = 0.326 x 20.00 x 450 x 300
= 880.88 kN
Puy1 = Puc + Pus(Total)
= 880.88 + (219.56)
= 1100.43 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 272/300
= 0.377
Muc = 880.88 x (0.5 x 300 - 0.377 x 300) = 32.50 kN-m
Muy1 = Muc + Mus(Total)
= 32.50 + (17.65)
= 50.15 kN-m
Pu/Puz = 0.706
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.706, an = 1.843
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((22.75/69.58)1.843) + ((41.28/50.15)1.843)
= 0.826 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG52 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1647.09 kN
MomentX,(Mx) = 7.25 kN-m
MomentY,(My) = 39.79 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C600.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 34.095 kN.m
My_MinEccen = 32.942 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 34.095 kN.m
My = max(My,My_MinEccen) + MuaddY = 39.793 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 7.25/1647.09 = 4 mm
Actual eccenY = My / P = 39.79/1647.09 = 24 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(4,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(24,20) = 24 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(24 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 14 nos. (2815 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00312 354.93 8.92 346.01 139.14 177 24.63
2 402 0.00265 348.91 8.92 339.99 136.72 118 16.13
3 402 0.00218 334.27 8.92 325.35 130.83 59 7.72
4 402 0.00171 311.57 8.73 302.84 121.78 0 0.00
5 402 0.00124 247.62 7.63 239.99 96.51 -59 -5.69
6 402 0.00077 153.64 5.54 148.10 59.56 -118 -7.03
7 402 0.00030 59.66 2.46 57.20 23.00 -177 -4.07
Total707.53 31.69
xux = 439 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 439/450 = 0.352
Puc = 0.352 x 20.00 x 300 x 450
= 949.22 kN
Pux1 = Puc + Pus(Total)
= 949.22 + (707.53)
= 1656.75 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 439/450
= 0.406
Muc = 949.22 x (0.5 x 450 - 0.406 x 450) = 40.05 kN-m
Mux1 = Muc + Mus(Total)
= 40.05 + (31.69)
= 71.73 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1407 0.00289 352.95 8.92 344.03 484.20 102 49.39
2 1407 0.00063 126.93 4.76 122.17 171.95 -102 -17.54
Total656.14 31.85
xuy = 309 mm               Puc = C1.fck.B.D
ku = 1.031
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.027
C1 = 0.446 x (1 - C3/6) = 0.370
Puc = 0.370 x 20.00 x 450 x 300
= 998.00 kN
Puy1 = Puc + Pus(Total)
= 998.00 + (656.14)
= 1654.14 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.426
Muc = 998.00 x (0.5 x 300 - 0.426 x 300) = 22.09 kN-m
Muy1 = Muc + Mus(Total)
= 22.09 + (31.85)
= 53.94 kN-m
Pu/Puz = 0.797
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.797, an = 1.996
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((34.09/71.73)1.996) + ((39.79/53.94)1.996)
= 0.772 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG52 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 14 nos. (2815 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 2198.18 kN
MomentX,(Mx) = 8.74 kN-m
MomentY,(My) = 37.12 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C600.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 45.502 kN.m
My_MinEccen = 43.964 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 45.502 kN.m
My = max(My,My_MinEccen) + MuaddY = 43.964 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 8.74/2198.18 = 4 mm
Actual eccenY = My / P = 37.12/2198.18 = 17 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(4,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(17,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 16 nos. (5027 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00310 354.76 8.92 345.84 217.30 175 38.03
2 628 0.00270 350.18 8.92 341.26 214.42 125 26.80
3 628 0.00230 338.60 8.92 329.68 207.14 75 15.54
4 628 0.00189 323.19 8.90 314.30 197.48 25 4.94
5 628 0.00149 293.71 8.35 285.36 179.30 -25 -4.48
6 628 0.00109 218.28 7.08 211.20 132.70 -75 -9.95
7 628 0.00069 137.99 5.09 132.90 83.50 -125 -10.44
8 628 0.00029 57.71 2.39 55.32 34.76 -175 -6.08
Total1266.60 54.35
xux = 436 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 436/450 = 0.349
Puc = 0.349 x 20.00 x 300 x 450
= 941.63 kN
Pux1 = Puc + Pus(Total)
= 941.63 + (1266.60)
= 2208.23 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 436/450
= 0.403
Muc = 941.63 x (0.5 x 450 - 0.403 x 450) = 41.10 kN-m
Mux1 = Muc + Mus(Total)
= 41.10 + (54.35)
= 95.45 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 2513 0.00285 352.56 8.92 343.64 863.67 100 86.37
2 2513 0.00069 138.15 5.10 133.05 334.39 -100 -33.44
Total1198.06 52.93
xuy = 314 mm               Puc = C1.fck.B.D
ku = 1.047
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.976
C1 = 0.446 x (1 - C3/6) = 0.373
Puc = 0.373 x 20.00 x 450 x 300
= 1008.29 kN
Puy1 = Puc + Pus(Total)
= 1008.29 + (1198.06)
= 2206.35 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.431
Muc = 1008.29 x (0.5 x 300 - 0.431 x 300) = 20.99 kN-m
Muy1 = Muc + Mus(Total)
= 20.99 + (52.93)
= 73.92 kN-m
Pu/Puz = 0.804
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.804, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((45.50/95.45)2.000) + ((43.96/73.92)2.000)
= 0.581 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 20 = 320 mm
               iii) 300 mm
               Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG52 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 16 nos. (5027 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 2764.57 kN
MomentX,(Mx) = 5.60 kN-m
MomentY,(My) = 22.64 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C600.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 60.129 kN.m
My_MinEccen = 55.291 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 60.129 kN.m
My = max(My,My_MinEccen) + MuaddY = 55.291 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 5.60/2764.57 = 2 mm
Actual eccenY = My / P = 22.64/2764.57 = 8 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(2,22) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(8,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #25 - 14 nos. (6872 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 982 0.00306 354.45 8.92 345.53 339.22 173 58.52
2 982 0.00263 348.39 8.92 339.47 333.27 115 38.33
3 982 0.00219 334.79 8.92 325.87 319.92 58 18.40
4 982 0.00176 314.60 8.79 305.82 300.24 0 0.00
5 982 0.00132 264.27 7.89 256.37 251.69 -58 -14.47
6 982 0.00089 177.20 6.15 171.05 167.93 -115 -19.31
7 982 0.00045 90.14 3.57 86.57 84.99 -173 -14.66
Total1797.27 66.79
xux = 457 mm               Puc = C1.fck.B.D
ku = 1.016
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.083
C1 = 0.446 x (1 - C3/6) = 0.366
Puc = 0.366 x 20.00 x 300 x 450
= 986.88 kN
Pux1 = Puc + Pus(Total)
= 986.88 + (1797.27)
= 2784.14 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.421
Muc = 986.88 x (0.5 x 450 - 0.421 x 450) = 34.93 kN-m
Mux1 = Muc + Mus(Total)
= 34.93 + (66.79)
= 101.72 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 3436 0.00274 351.41 8.92 342.49 1176.84 98 114.74
2 3436 0.00084 167.08 5.90 161.19 553.86 -98 -54.00
Total1730.70 60.74
xuy = 333 mm               Puc = C1.fck.B.D
ku = 1.109
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.805
C1 = 0.446 x (1 - C3/6) = 0.386
Puc = 0.386 x 20.00 x 450 x 300
= 1042.61 kN
Puy1 = Puc + Pus(Total)
= 1042.61 + (1730.70)
= 2773.31 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.445
Muc = 1042.61 x (0.5 x 300 - 0.445 x 300) = 17.31 kN-m
Muy1 = Muc + Mus(Total)
= 17.31 + (60.74)
= 78.05 kN-m
Pu/Puz = 0.840
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.840, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((60.13/101.72)2.000) + ((55.29/78.05)2.000)
= 0.851 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 25/4 = 6 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (6, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 25 = 400 mm
               iii) 300 mm
               Required spacing = minimum of (300, 400, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG52 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #25 - 14 nos. (6872 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG53 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 460.41 kN
MomentX,(Mx) = 1.79 kN-m
MomentY,(My) = 14.05 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C610.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 9.531 kN.m
My_MinEccen = 9.208 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 9.531 kN.m
My = max(My,My_MinEccen) + MuaddY = 14.053 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 1.79/460.41 = 4 mm
Actual eccenY = My / P = 14.05/460.41 = 31 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(4,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(31,20) = 31 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(31 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00277 351.86 8.92 342.94 77.57 179 13.89
2 226 0.00134 268.33 7.95 260.37 58.90 90 5.27
3 226 -0.00008 -16.80 0.00 -16.80 -3.80 0 0.00
4 226 -0.00151 -295.30 0.00 -295.30 -66.79 -90 5.98
5 226 -0.00294 -353.33 0.00 -353.33 -79.92 -179 14.31
Total-14.05 39.44
xux = 220 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 220/450 = 0.176
Puc = 0.176 x 20.00 x 300 x 450
= 474.61 kN
Pux1 = Puc + Pus(Total)
= 474.61 + (-14.05)
= 460.56 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 220/450
= 0.203
Muc = 474.61 x (0.5 x 450 - 0.203 x 450) = 63.40 kN-m
Mux1 = Muc + Mus(Total)
= 63.40 + (39.44)
= 102.85 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00239 342.14 8.92 333.22 188.43 104 19.60
2 565 -0.00262 -348.14 0.00 -348.14 -196.87 -104 20.47
Total-8.44 40.07
xuy = 145 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 145/300 = 0.174
Puc = 0.174 x 20.00 x 450 x 300
= 470.81 kN
Puy1 = Puc + Pus(Total)
= 470.81 + (-8.44)
= 462.37 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 145/300
= 0.201
Muc = 470.81 x (0.5 x 300 - 0.201 x 300) = 42.16 kN-m
Muy1 = Muc + Mus(Total)
= 42.16 + (40.07)
= 82.23 kN-m
Pu/Puz = 0.296
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.296, an = 1.160
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((9.53/102.85)1.160) + ((14.05/82.23)1.160)
= 0.192 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG53 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 957.53 kN
MomentX,(Mx) = 2.06 kN-m
MomentY,(My) = 12.92 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C610.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 19.821 kN.m
My_MinEccen = 19.151 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 19.821 kN.m
My = max(My,My_MinEccen) + MuaddY = 19.151 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 2.06/957.53 = 2 mm
Actual eccenY = My / P = 12.92/957.53 = 13 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(2,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(13,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00304 354.27 8.92 345.35 78.12 179 13.98
2 226 0.00215 333.29 8.92 324.37 73.37 90 6.57
3 226 0.00126 252.00 7.70 244.30 55.26 0 0.00
4 226 0.00037 73.80 2.99 70.81 16.02 -90 -1.43
5 226 -0.00052 -104.41 0.00 -104.41 -23.62 -179 4.23
Total199.15 23.34
xux = 352 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 352/450 = 0.281
Puc = 0.281 x 20.00 x 300 x 450
= 759.38 kN
Pux1 = Puc + Pus(Total)
= 759.38 + (199.15)
= 958.52 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 352/450
= 0.325
Muc = 759.38 x (0.5 x 450 - 0.325 x 450) = 59.80 kN-m
Mux1 = Muc + Mus(Total)
= 59.80 + (23.34)
= 83.14 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00284 352.47 8.92 343.55 194.27 104 20.20
2 565 -0.00016 -32.96 0.00 -32.96 -18.64 -104 1.94
Total175.63 22.14
xuy = 243 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 243/300 = 0.291
Puc = 0.291 x 20.00 x 450 x 300
= 785.95 kN
Puy1 = Puc + Pus(Total)
= 785.95 + (175.63)
= 961.59 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 243/300
= 0.336
Muc = 785.95 x (0.5 x 300 - 0.336 x 300) = 38.58 kN-m
Muy1 = Muc + Mus(Total)
= 38.58 + (22.14)
= 60.72 kN-m
Pu/Puz = 0.615
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.615, an = 1.692
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((19.82/83.14)1.692) + ((19.15/60.72)1.692)
= 0.230 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG53 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1410.77 kN
MomentX,(Mx) = 2.83 kN-m
MomentY,(My) = 10.90 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C610.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 29.203 kN.m
My_MinEccen = 28.215 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 29.203 kN.m
My = max(My,My_MinEccen) + MuaddY = 28.215 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 2.83/1410.77 = 2 mm
Actual eccenY = My / P = 10.90/1410.77 = 8 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(2,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(8,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 8 nos. (1608 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00304 354.26 8.92 345.34 138.87 177 24.58
2 402 0.00219 334.83 8.92 325.91 131.06 59 7.73
3 402 0.00134 268.97 7.96 261.01 104.96 -59 -6.19
4 402 0.00050 99.33 3.88 95.45 38.38 -177 -6.79
Total413.27 19.33
xux = 471 mm               Puc = C1.fck.B.D
ku = 1.047
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.976
C1 = 0.446 x (1 - C3/6) = 0.373
Puc = 0.373 x 20.00 x 300 x 450
= 1008.29 kN
Pux1 = Puc + Pus(Total)
= 1008.29 + (413.27)
= 1421.56 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.431
Muc = 1008.29 x (0.5 x 450 - 0.431 x 450) = 31.49 kN-m
Mux1 = Muc + Mus(Total)
= 31.49 + (19.33)
= 50.81 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 804 0.00283 352.39 8.92 343.47 276.23 102 28.18
2 804 0.00073 146.64 5.34 141.30 113.64 -102 -11.59
Total389.87 16.58
xuy = 323 mm               Puc = C1.fck.B.D
ku = 1.078
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.884
C1 = 0.446 x (1 - C3/6) = 0.380
Puc = 0.380 x 20.00 x 450 x 300
= 1026.69 kN
Puy1 = Puc + Pus(Total)
= 1026.69 + (389.87)
= 1416.56 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.438
Muc = 1026.69 x (0.5 x 300 - 0.438 x 300) = 19.02 kN-m
Muy1 = Muc + Mus(Total)
= 19.02 + (16.58)
= 35.60 kN-m
Pu/Puz = 0.829
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.829, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((29.20/50.81)2.000) + ((28.22/35.60)2.000)
= 0.958 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG53 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 8 nos. (1608 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1790.27 kN
MomentX,(Mx) = 4.49 kN-m
MomentY,(My) = 7.59 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C610.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 37.059 kN.m
My_MinEccen = 35.805 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 37.059 kN.m
My = max(My,My_MinEccen) + MuaddY = 35.805 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 4.49/1790.27 = 3 mm
Actual eccenY = My / P = 7.59/1790.27 = 4 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(3,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(4,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 16 nos. (3217 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00313 355.01 8.92 346.09 139.17 177 24.63
2 402 0.00273 351.11 8.92 342.19 137.60 126 17.40
3 402 0.00234 340.23 8.92 331.31 133.23 76 10.11
4 402 0.00195 325.78 8.91 316.87 127.42 25 3.22
5 402 0.00155 299.44 8.48 290.96 117.00 -25 -2.96
6 402 0.00116 232.00 7.35 224.65 90.34 -76 -6.85
7 402 0.00077 153.33 5.53 147.81 59.44 -126 -7.51
8 402 0.00037 74.67 3.02 71.65 28.81 -177 -5.10
Total833.01 32.93
xux = 450 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 450/450 = 0.360
Puc = 0.360 x 20.00 x 300 x 450
= 972.00 kN
Pux1 = Puc + Pus(Total)
= 972.00 + (833.01)
= 1805.01 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 450/450
= 0.416
Muc = 972.00 x (0.5 x 450 - 0.416 x 450) = 36.74 kN-m
Mux1 = Muc + Mus(Total)
= 36.74 + (32.93)
= 69.67 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1608 0.00283 352.39 8.92 343.47 552.46 102 56.35
2 1608 0.00073 146.64 5.34 141.30 227.28 -102 -23.18
Total779.74 33.17
xuy = 323 mm               Puc = C1.fck.B.D
ku = 1.078
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.884
C1 = 0.446 x (1 - C3/6) = 0.380
Puc = 0.380 x 20.00 x 450 x 300
= 1026.69 kN
Puy1 = Puc + Pus(Total)
= 1026.69 + (779.74)
= 1806.43 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.438
Muc = 1026.69 x (0.5 x 300 - 0.438 x 300) = 19.02 kN-m
Muy1 = Muc + Mus(Total)
= 19.02 + (33.17)
= 52.19 kN-m
Pu/Puz = 0.818
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.818, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((37.06/69.67)2.000) + ((35.81/52.19)2.000)
= 0.754 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG53 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 16 nos. (3217 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 2029.58 kN
MomentX,(Mx) = 4.39 kN-m
MomentY,(My) = 2.49 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C610.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 44.143 kN.m
My_MinEccen = 40.592 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 44.143 kN.m
My = max(My,My_MinEccen) + MuaddY = 40.592 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 4.39/2029.58 = 2 mm
Actual eccenY = My / P = 2.49/2029.58 = 1 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(2,22) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(1,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 14 nos. (4398 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00310 354.76 8.92 345.84 217.30 175 38.03
2 628 0.00263 348.46 8.92 339.54 213.34 117 24.89
3 628 0.00216 333.69 8.92 324.77 204.06 58 11.90
4 628 0.00169 310.67 8.71 301.96 189.72 0 0.00
5 628 0.00123 245.04 7.58 237.46 149.20 -58 -8.70
6 628 0.00076 151.37 5.47 145.90 91.67 -117 -10.70
7 628 0.00029 57.71 2.39 55.32 34.76 -175 -6.08
Total1100.05 49.34
xux = 436 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 436/450 = 0.349
Puc = 0.349 x 20.00 x 300 x 450
= 941.63 kN
Pux1 = Puc + Pus(Total)
= 941.63 + (1100.05)
= 2041.68 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 436/450
= 0.403
Muc = 941.63 x (0.5 x 450 - 0.403 x 450) = 41.10 kN-m
Mux1 = Muc + Mus(Total)
= 41.10 + (49.34)
= 90.44 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 2199 0.00287 352.75 8.92 343.83 756.13 100 75.61
2 2199 0.00066 131.36 4.90 126.46 278.10 -100 -27.81
Total1034.24 47.80
xuy = 309 mm               Puc = C1.fck.B.D
ku = 1.031
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.027
C1 = 0.446 x (1 - C3/6) = 0.370
Puc = 0.370 x 20.00 x 450 x 300
= 998.00 kN
Puy1 = Puc + Pus(Total)
= 998.00 + (1034.24)
= 2032.23 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.426
Muc = 998.00 x (0.5 x 300 - 0.426 x 300) = 22.09 kN-m
Muy1 = Muc + Mus(Total)
= 22.09 + (47.80)
= 69.90 kN-m
Pu/Puz = 0.798
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.798, an = 1.996
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((44.14/90.44)1.996) + ((40.59/69.90)1.996)
= 0.577 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 20 = 320 mm
               iii) 300 mm
               Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG53 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 14 nos. (4398 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG54 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 430.91 kN
MomentX,(Mx) = 5.40 kN-m
MomentY,(My) = 0.25 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C620.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 8.920 kN.m
My_MinEccen = 8.618 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 8.920 kN.m
My = max(My,My_MinEccen) + MuaddY = 8.618 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 5.40/430.91 = 13 mm
Actual eccenY = My / P = 0.25/430.91 = 1 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(13,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(1,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00274 351.36 8.92 342.44 77.46 179 13.87
2 226 0.00127 254.06 7.73 246.32 55.72 90 4.99
3 226 -0.00020 -40.50 0.00 -40.50 -9.16 0 0.00
4 226 -0.00168 -309.52 0.00 -309.52 -70.01 -90 6.27
5 226 -0.00315 -355.20 0.00 -355.20 -80.34 -179 14.38
Total-26.34 39.50
xux = 213 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 213/450 = 0.170
Puc = 0.170 x 20.00 x 300 x 450
= 459.42 kN
Pux1 = Puc + Pus(Total)
= 459.42 + (-26.34)
= 433.08 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 213/450
= 0.197
Muc = 459.42 x (0.5 x 450 - 0.197 x 450) = 62.72 kN-m
Mux1 = Muc + Mus(Total)
= 62.72 + (39.50)
= 102.22 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00233 339.71 8.92 330.79 187.05 104 19.45
2 565 -0.00298 -353.76 0.00 -353.76 -200.05 -104 20.80
Total-12.99 40.26
xuy = 137 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 137/300 = 0.165
Puc = 0.165 x 20.00 x 450 x 300
= 444.23 kN
Puy1 = Puc + Pus(Total)
= 444.23 + (-12.99)
= 431.24 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 137/300
= 0.190
Muc = 444.23 x (0.5 x 300 - 0.190 x 300) = 41.30 kN-m
Muy1 = Muc + Mus(Total)
= 41.30 + (40.26)
= 81.56 kN-m
Pu/Puz = 0.277
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.277, an = 1.128
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((8.92/102.22)1.128) + ((8.62/81.56)1.128)
= 0.143 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG54 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 888.39 kN
MomentX,(Mx) = 0.66 kN-m
MomentY,(My) = 0.17 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C620.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 18.390 kN.m
My_MinEccen = 17.768 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 18.390 kN.m
My = max(My,My_MinEccen) + MuaddY = 17.768 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 0.66/888.39 = 1 mm
Actual eccenY = My / P = 0.17/888.39 = 0 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(1,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(0,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00302 354.03 8.92 345.11 78.06 179 13.97
2 226 0.00207 330.40 8.92 321.48 72.72 90 6.51
3 226 0.00113 225.93 7.23 218.70 49.47 0 0.00
4 226 0.00019 37.35 1.59 35.76 8.09 -90 -0.72
5 226 -0.00076 -151.23 0.00 -151.23 -34.21 -179 6.12
Total174.13 25.88
xux = 332 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 332/450 = 0.266
Puc = 0.266 x 20.00 x 300 x 450
= 717.61 kN
Pux1 = Puc + Pus(Total)
= 717.61 + (174.13)
= 891.74 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 332/450
= 0.307
Muc = 717.61 x (0.5 x 450 - 0.307 x 450) = 62.28 kN-m
Mux1 = Muc + Mus(Total)
= 62.28 + (25.88)
= 88.16 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00280 352.14 8.92 343.22 194.09 104 20.19
2 565 -0.00037 -74.10 0.00 -74.10 -41.90 -104 4.36
Total152.19 24.54
xuy = 230 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 230/300 = 0.276
Puc = 0.276 x 20.00 x 450 x 300
= 744.19 kN
Puy1 = Puc + Pus(Total)
= 744.19 + (152.19)
= 896.37 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 230/300
= 0.319
Muc = 744.19 x (0.5 x 300 - 0.319 x 300) = 40.52 kN-m
Muy1 = Muc + Mus(Total)
= 40.52 + (24.54)
= 65.06 kN-m
Pu/Puz = 0.571
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.571, an = 1.618
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((18.39/88.16)1.618) + ((17.77/65.06)1.618)
= 0.202 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG54 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1329.18 kN
MomentX,(Mx) = 1.84 kN-m
MomentY,(My) = 0.89 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C620.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 27.514 kN.m
My_MinEccen = 26.584 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 27.514 kN.m
My = max(My,My_MinEccen) + MuaddY = 26.584 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 1.84/1329.18 = 1 mm
Actual eccenY = My / P = 0.89/1329.18 = 1 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(1,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(1,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 12 nos. (1357 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00311 354.88 8.92 345.96 78.25 179 14.01
2 226 0.00257 346.91 8.92 337.99 76.45 107 8.21
3 226 0.00203 328.76 8.92 319.84 72.35 36 2.59
4 226 0.00149 293.02 8.33 284.69 64.40 -36 -2.31
5 226 0.00094 188.71 6.43 182.28 41.23 -107 -4.43
6 226 0.00040 80.30 3.22 77.08 17.43 -179 -3.12
Total350.11 14.95
xux = 457 mm               Puc = C1.fck.B.D
ku = 1.016
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.083
C1 = 0.446 x (1 - C3/6) = 0.366
Puc = 0.366 x 20.00 x 300 x 450
= 986.88 kN
Pux1 = Puc + Pus(Total)
= 986.88 + (350.11)
= 1336.99 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.421
Muc = 986.88 x (0.5 x 450 - 0.421 x 450) = 34.93 kN-m
Mux1 = Muc + Mus(Total)
= 34.93 + (14.95)
= 49.88 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 679 0.00287 352.75 8.92 343.83 233.32 104 24.26
2 679 0.00068 136.19 5.04 131.15 88.99 -104 -9.26
Total322.31 15.01
xuy = 319 mm               Puc = C1.fck.B.D
ku = 1.063
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.929
C1 = 0.446 x (1 - C3/6) = 0.377
Puc = 0.377 x 20.00 x 450 x 300
= 1017.83 kN
Puy1 = Puc + Pus(Total)
= 1017.83 + (322.31)
= 1340.14 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.435
Muc = 1017.83 x (0.5 x 300 - 0.435 x 300) = 19.97 kN-m
Muy1 = Muc + Mus(Total)
= 19.97 + (15.01)
= 34.98 kN-m
Pu/Puz = 0.818
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.818, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((27.51/49.88)2.000) + ((26.58/34.98)2.000)
= 0.882 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG54 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 12 nos. (1357 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1744.34 kN
MomentX,(Mx) = 4.52 kN-m
MomentY,(My) = 2.65 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C620.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 36.108 kN.m
My_MinEccen = 34.887 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 36.108 kN.m
My = max(My,My_MinEccen) + MuaddY = 34.887 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 4.52/1744.34 = 3 mm
Actual eccenY = My / P = 2.65/1744.34 = 2 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(3,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(2,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 16 nos. (3217 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00311 354.90 8.92 345.98 139.13 177 24.63
2 402 0.00271 350.48 8.92 341.56 137.35 126 17.36
3 402 0.00230 338.85 8.92 329.93 132.67 76 10.06
4 402 0.00190 323.34 8.90 314.44 126.44 25 3.20
5 402 0.00149 293.49 8.34 285.15 114.66 -25 -2.90
6 402 0.00108 216.90 7.05 209.85 84.39 -76 -6.40
7 402 0.00068 135.70 5.03 130.67 52.55 -126 -6.64
8 402 0.00027 54.49 2.26 52.23 21.00 -177 -3.72
Total808.20 35.59
xux = 436 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 436/450 = 0.349
Puc = 0.349 x 20.00 x 300 x 450
= 941.63 kN
Pux1 = Puc + Pus(Total)
= 941.63 + (808.20)
= 1749.82 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 436/450
= 0.403
Muc = 941.63 x (0.5 x 450 - 0.403 x 450) = 41.10 kN-m
Mux1 = Muc + Mus(Total)
= 41.10 + (35.59)
= 76.69 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1608 0.00289 352.95 8.92 344.03 553.37 102 56.44
2 1608 0.00063 126.93 4.76 122.17 196.51 -102 -20.04
Total749.88 36.40
xuy = 309 mm               Puc = C1.fck.B.D
ku = 1.031
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.027
C1 = 0.446 x (1 - C3/6) = 0.370
Puc = 0.370 x 20.00 x 450 x 300
= 998.00 kN
Puy1 = Puc + Pus(Total)
= 998.00 + (749.88)
= 1747.88 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.426
Muc = 998.00 x (0.5 x 300 - 0.426 x 300) = 22.09 kN-m
Muy1 = Muc + Mus(Total)
= 22.09 + (36.40)
= 58.49 kN-m
Pu/Puz = 0.797
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.797, an = 1.996
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((36.11/76.69)1.996) + ((34.89/58.49)1.996)
= 0.579 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG54 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 16 nos. (3217 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 2121.14 kN
MomentX,(Mx) = 7.57 kN-m
MomentY,(My) = 2.08 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C620.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 46.135 kN.m
My_MinEccen = 42.423 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 46.135 kN.m
My = max(My,My_MinEccen) + MuaddY = 42.423 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 7.57/2121.14 = 4 mm
Actual eccenY = My / P = 2.08/2121.14 = 1 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(4,22) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(1,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 14 nos. (4398 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00308 354.61 8.92 345.69 217.21 175 38.01
2 628 0.00264 348.71 8.92 339.79 213.50 117 24.91
3 628 0.00220 335.02 8.92 326.10 204.90 58 11.95
4 628 0.00176 314.60 8.79 305.82 192.15 0 0.00
5 628 0.00132 263.01 7.87 255.13 160.30 -58 -9.35
6 628 0.00087 174.68 6.09 168.59 105.93 -117 -12.36
7 628 0.00043 86.35 3.44 82.92 52.10 -175 -9.12
Total1146.08 44.04
xux = 457 mm               Puc = C1.fck.B.D
ku = 1.016
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.083
C1 = 0.446 x (1 - C3/6) = 0.366
Puc = 0.366 x 20.00 x 300 x 450
= 986.88 kN
Pux1 = Puc + Pus(Total)
= 986.88 + (1146.08)
= 2132.96 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.421
Muc = 986.88 x (0.5 x 450 - 0.421 x 450) = 34.93 kN-m
Mux1 = Muc + Mus(Total)
= 34.93 + (44.04)
= 78.97 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 2199 0.00279 352.04 8.92 343.12 754.56 100 75.46
2 2199 0.00078 156.60 5.62 150.98 332.03 -100 -33.20
Total1086.59 42.25
xuy = 328 mm               Puc = C1.fck.B.D
ku = 1.094
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.843
C1 = 0.446 x (1 - C3/6) = 0.383
Puc = 0.383 x 20.00 x 450 x 300
= 1034.93 kN
Puy1 = Puc + Pus(Total)
= 1034.93 + (1086.59)
= 2121.52 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.442
Muc = 1034.93 x (0.5 x 300 - 0.442 x 300) = 18.14 kN-m
Muy1 = Muc + Mus(Total)
= 18.14 + (42.25)
= 60.39 kN-m
Pu/Puz = 0.834
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.834, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((46.13/78.97)2.000) + ((42.42/60.39)2.000)
= 0.835 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 20 = 320 mm
               iii) 300 mm
               Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG54 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 14 nos. (4398 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG55 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 459.19 kN
MomentX,(Mx) = 11.11 kN-m
MomentY,(My) = 5.79 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C630.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 9.505 kN.m
My_MinEccen = 9.184 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 11.114 kN.m
My = max(My,My_MinEccen) + MuaddY = 9.184 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 11.11/459.19 = 24 mm
Actual eccenY = My / P = 5.79/459.19 = 13 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(24,21) = 24 mm
eccenY = max(Actual eccenY,eccenYMin) = max(13,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(24 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00277 351.86 8.92 342.94 77.57 179 13.89
2 226 0.00134 268.33 7.95 260.37 58.90 90 5.27
3 226 -0.00008 -16.80 0.00 -16.80 -3.80 0 0.00
4 226 -0.00151 -295.30 0.00 -295.30 -66.79 -90 5.98
5 226 -0.00294 -353.33 0.00 -353.33 -79.92 -179 14.31
Total-14.05 39.44
xux = 220 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 220/450 = 0.176
Puc = 0.176 x 20.00 x 300 x 450
= 474.61 kN
Pux1 = Puc + Pus(Total)
= 474.61 + (-14.05)
= 460.56 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 220/450
= 0.203
Muc = 474.61 x (0.5 x 450 - 0.203 x 450) = 63.40 kN-m
Mux1 = Muc + Mus(Total)
= 63.40 + (39.44)
= 102.85 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00239 342.14 8.92 333.22 188.43 104 19.60
2 565 -0.00262 -348.14 0.00 -348.14 -196.87 -104 20.47
Total-8.44 40.07
xuy = 145 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 145/300 = 0.174
Puc = 0.174 x 20.00 x 450 x 300
= 470.81 kN
Puy1 = Puc + Pus(Total)
= 470.81 + (-8.44)
= 462.37 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 145/300
= 0.201
Muc = 470.81 x (0.5 x 300 - 0.201 x 300) = 42.16 kN-m
Muy1 = Muc + Mus(Total)
= 42.16 + (40.07)
= 82.23 kN-m
Pu/Puz = 0.295
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.295, an = 1.158
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((11.11/102.85)1.158) + ((9.18/82.23)1.158)
= 0.155 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG55 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 937.23 kN
MomentX,(Mx) = 10.89 kN-m
MomentY,(My) = 4.33 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C630.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 19.401 kN.m
My_MinEccen = 18.745 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 19.401 kN.m
My = max(My,My_MinEccen) + MuaddY = 18.745 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 10.89/937.23 = 12 mm
Actual eccenY = My / P = 4.33/937.23 = 5 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(12,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(5,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00304 354.23 8.92 345.31 78.11 179 13.98
2 226 0.00214 332.79 8.92 323.87 73.26 90 6.56
3 226 0.00124 247.47 7.62 239.85 54.25 0 0.00
4 226 0.00034 67.47 2.76 64.71 14.64 -90 -1.31
5 226 -0.00056 -112.53 0.00 -112.53 -25.45 -179 4.56
Total194.80 23.78
xux = 348 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 348/450 = 0.278
Puc = 0.278 x 20.00 x 300 x 450
= 751.78 kN
Pux1 = Puc + Pus(Total)
= 751.78 + (194.80)
= 946.58 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 348/450
= 0.322
Muc = 751.78 x (0.5 x 450 - 0.322 x 450) = 60.30 kN-m
Mux1 = Muc + Mus(Total)
= 60.30 + (23.78)
= 84.09 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00283 352.38 8.92 343.46 194.22 104 20.20
2 565 -0.00022 -43.74 0.00 -43.74 -24.73 -104 2.57
Total169.49 22.77
xuy = 239 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 239/300 = 0.287
Puc = 0.287 x 20.00 x 450 x 300
= 774.56 kN
Puy1 = Puc + Pus(Total)
= 774.56 + (169.49)
= 944.05 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 239/300
= 0.331
Muc = 774.56 x (0.5 x 300 - 0.331 x 300) = 39.15 kN-m
Muy1 = Muc + Mus(Total)
= 39.15 + (22.77)
= 61.93 kN-m
Pu/Puz = 0.602
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.602, an = 1.670
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((19.40/84.09)1.670) + ((18.74/61.93)1.670)
= 0.222 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG55 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1385.10 kN
MomentX,(Mx) = 10.53 kN-m
MomentY,(My) = 2.49 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C630.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 28.672 kN.m
My_MinEccen = 27.702 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 28.672 kN.m
My = max(My,My_MinEccen) + MuaddY = 27.702 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 10.53/1385.10 = 8 mm
Actual eccenY = My / P = 2.49/1385.10 = 2 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(8,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(2,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 14 nos. (1583 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00311 354.88 8.92 345.96 78.25 179 14.01
2 226 0.00266 349.23 8.92 340.31 76.98 119 9.19
3 226 0.00221 335.39 8.92 326.47 73.85 60 4.41
4 226 0.00176 314.60 8.79 305.82 69.17 -0 -0.00
5 226 0.00130 260.99 7.84 253.14 57.26 -60 -3.42
6 226 0.00085 170.64 5.99 164.65 37.24 -119 -4.44
7 226 0.00040 80.30 3.22 77.08 17.43 -179 -3.12
Total410.19 16.62
xux = 457 mm               Puc = C1.fck.B.D
ku = 1.016
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.083
C1 = 0.446 x (1 - C3/6) = 0.366
Puc = 0.366 x 20.00 x 300 x 450
= 986.88 kN
Pux1 = Puc + Pus(Total)
= 986.88 + (410.19)
= 1397.07 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.421
Muc = 986.88 x (0.5 x 450 - 0.421 x 450) = 34.93 kN-m
Mux1 = Muc + Mus(Total)
= 34.93 + (16.62)
= 51.54 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 792 0.00287 352.75 8.92 343.83 272.20 104 28.31
2 792 0.00068 136.19 5.04 131.15 103.83 -104 -10.80
Total376.03 17.51
xuy = 319 mm               Puc = C1.fck.B.D
ku = 1.063
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.929
C1 = 0.446 x (1 - C3/6) = 0.377
Puc = 0.377 x 20.00 x 450 x 300
= 1017.83 kN
Puy1 = Puc + Pus(Total)
= 1017.83 + (376.03)
= 1393.86 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.435
Muc = 1017.83 x (0.5 x 300 - 0.435 x 300) = 19.97 kN-m
Muy1 = Muc + Mus(Total)
= 19.97 + (17.51)
= 37.48 kN-m
Pu/Puz = 0.818
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.818, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((28.67/51.54)2.000) + ((27.70/37.48)2.000)
= 0.856 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG55 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 14 nos. (1583 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1782.95 kN
MomentX,(Mx) = 7.18 kN-m
MomentY,(My) = 0.97 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C630.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 36.907 kN.m
My_MinEccen = 35.659 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 36.907 kN.m
My = max(My,My_MinEccen) + MuaddY = 35.659 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 7.18/1782.95 = 4 mm
Actual eccenY = My / P = 0.97/1782.95 = 1 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(4,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(1,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 16 nos. (3217 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00312 354.98 8.92 346.06 139.16 177 24.63
2 402 0.00273 350.96 8.92 342.04 137.54 126 17.39
3 402 0.00233 339.89 8.92 330.97 133.09 76 10.10
4 402 0.00193 325.33 8.91 316.42 127.24 25 3.22
5 402 0.00154 297.98 8.44 289.54 116.43 -25 -2.94
6 402 0.00114 228.31 7.28 221.04 88.88 -76 -6.74
7 402 0.00075 149.03 5.41 143.62 57.75 -126 -7.30
8 402 0.00035 69.74 2.84 66.90 26.90 -177 -4.76
Total827.01 33.58
xux = 446 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 446/450 = 0.357
Puc = 0.357 x 20.00 x 300 x 450
= 964.41 kN
Pux1 = Puc + Pus(Total)
= 964.41 + (827.01)
= 1791.41 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 446/450
= 0.413
Muc = 964.41 x (0.5 x 450 - 0.413 x 450) = 37.87 kN-m
Mux1 = Muc + Mus(Total)
= 37.87 + (33.58)
= 71.45 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1608 0.00285 352.56 8.92 343.64 552.75 102 56.38
2 1608 0.00070 140.39 5.16 135.23 217.52 -102 -22.19
Total770.27 34.19
xuy = 319 mm               Puc = C1.fck.B.D
ku = 1.063
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.929
C1 = 0.446 x (1 - C3/6) = 0.377
Puc = 0.377 x 20.00 x 450 x 300
= 1017.83 kN
Puy1 = Puc + Pus(Total)
= 1017.83 + (770.27)
= 1788.10 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.435
Muc = 1017.83 x (0.5 x 300 - 0.435 x 300) = 19.97 kN-m
Muy1 = Muc + Mus(Total)
= 19.97 + (34.19)
= 54.16 kN-m
Pu/Puz = 0.815
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.815, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((36.91/71.45)2.000) + ((35.66/54.16)2.000)
= 0.700 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG55 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 16 nos. (3217 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 2092.98 kN
MomentX,(Mx) = 2.96 kN-m
MomentY,(My) = 1.53 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C630.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 45.522 kN.m
My_MinEccen = 41.860 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 45.522 kN.m
My = max(My,My_MinEccen) + MuaddY = 41.860 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 2.96/2092.98 = 1 mm
Actual eccenY = My / P = 1.53/2092.98 = 1 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(1,22) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(1,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 14 nos. (4398 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00311 354.87 8.92 345.95 217.37 175 38.04
2 628 0.00266 349.16 8.92 340.24 213.78 117 24.94
3 628 0.00220 335.22 8.92 326.30 205.02 58 11.96
4 628 0.00175 314.19 8.78 305.41 191.89 0 0.00
5 628 0.00130 259.26 7.82 251.44 157.99 -58 -9.22
6 628 0.00084 168.52 5.93 162.59 102.16 -117 -11.92
7 628 0.00039 77.78 3.13 74.65 46.90 -175 -8.21
Total1135.11 45.60
xux = 450 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 450/450 = 0.360
Puc = 0.360 x 20.00 x 300 x 450
= 972.00 kN
Pux1 = Puc + Pus(Total)
= 972.00 + (1135.11)
= 2107.11 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 450/450
= 0.416
Muc = 972.00 x (0.5 x 450 - 0.416 x 450) = 36.74 kN-m
Mux1 = Muc + Mus(Total)
= 36.74 + (45.60)
= 82.34 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 2199 0.00281 352.21 8.92 343.29 754.93 100 75.49
2 2199 0.00075 150.74 5.46 145.29 319.51 -100 -31.95
Total1074.43 43.54
xuy = 323 mm               Puc = C1.fck.B.D
ku = 1.078
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.884
C1 = 0.446 x (1 - C3/6) = 0.380
Puc = 0.380 x 20.00 x 450 x 300
= 1026.69 kN
Puy1 = Puc + Pus(Total)
= 1026.69 + (1074.43)
= 2101.12 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.438
Muc = 1026.69 x (0.5 x 300 - 0.438 x 300) = 19.02 kN-m
Muy1 = Muc + Mus(Total)
= 19.02 + (43.54)
= 62.56 kN-m
Pu/Puz = 0.823
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.823, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((45.52/82.34)2.000) + ((41.86/62.56)2.000)
= 0.753 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 20 = 320 mm
               iii) 300 mm
               Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG55 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 14 nos. (4398 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG56 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 513.45 kN
MomentX,(Mx) = 14.31 kN-m
MomentY,(My) = 27.32 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C640.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 10.628 kN.m
My_MinEccen = 10.269 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 14.307 kN.m
My = max(My,My_MinEccen) + MuaddY = 27.320 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 14.31/513.45 = 28 mm
Actual eccenY = My / P = 27.32/513.45 = 53 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(28,21) = 28 mm
eccenY = max(Actual eccenY,eccenYMin) = max(53,20) = 53 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(28 mm) > 0.05 x 300(15 mm)
and eccenY(53 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00281 352.23 8.92 343.31 77.65 179 13.90
2 226 0.00146 290.96 8.28 282.68 63.94 90 5.72
3 226 0.00012 23.77 1.03 22.74 5.14 0 0.00
4 226 -0.00123 -245.21 0.00 -245.21 -55.47 -90 4.96
5 226 -0.00257 -346.94 0.00 -346.94 -78.48 -179 14.05
Total12.80 38.63
xux = 233 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 233/450 = 0.186
Puc = 0.186 x 20.00 x 300 x 450
= 503.09 kN
Pux1 = Puc + Pus(Total)
= 503.09 + (12.80)
= 515.88 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 233/450
= 0.215
Muc = 503.09 x (0.5 x 450 - 0.215 x 450) = 64.45 kN-m
Mux1 = Muc + Mus(Total)
= 64.45 + (38.63)
= 103.08 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00248 344.66 8.92 335.74 189.86 104 19.75
2 565 -0.00212 -332.12 0.00 -332.12 -187.81 -104 19.53
Total2.04 39.28
xuy = 158 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 158/300 = 0.190
Puc = 0.190 x 20.00 x 450 x 300
= 512.58 kN
Puy1 = Puc + Pus(Total)
= 512.58 + (2.04)
= 514.62 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 158/300
= 0.219
Muc = 512.58 x (0.5 x 300 - 0.219 x 300) = 43.15 kN-m
Muy1 = Muc + Mus(Total)
= 43.15 + (39.28)
= 82.43 kN-m
Pu/Puz = 0.330
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.330, an = 1.216
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((14.31/103.08)1.216) + ((27.32/82.43)1.216)
= 0.352 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG56 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1030.12 kN
MomentX,(Mx) = 12.63 kN-m
MomentY,(My) = 25.26 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C640.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 21.323 kN.m
My_MinEccen = 20.602 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 21.323 kN.m
My = max(My,My_MinEccen) + MuaddY = 25.264 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 12.63/1030.12 = 12 mm
Actual eccenY = My / P = 25.26/1030.12 = 25 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(12,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(25,20) = 25 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(25 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00307 354.53 8.92 345.61 78.18 179 13.99
2 226 0.00224 336.53 8.92 327.61 74.10 90 6.63
3 226 0.00141 281.31 8.13 273.17 61.79 0 0.00
4 226 0.00057 114.76 4.38 110.38 24.97 -90 -2.23
5 226 -0.00026 -51.78 0.00 -51.78 -11.71 -179 2.10
Total227.32 20.49
xux = 376 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 376/450 = 0.301
Puc = 0.301 x 20.00 x 300 x 450
= 812.53 kN
Pux1 = Puc + Pus(Total)
= 812.53 + (227.32)
= 1039.85 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 376/450
= 0.348
Muc = 812.53 x (0.5 x 450 - 0.348 x 450) = 55.67 kN-m
Mux1 = Muc + Mus(Total)
= 55.67 + (20.49)
= 76.16 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00288 352.81 8.92 343.89 194.47 104 20.22
2 565 0.00005 10.35 0.46 9.90 5.60 -104 -0.58
Total200.06 19.64
xuy = 258 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 258/300 = 0.309
Puc = 0.309 x 20.00 x 450 x 300
= 835.31 kN
Puy1 = Puc + Pus(Total)
= 835.31 + (200.06)
= 1035.37 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 258/300
= 0.358
Muc = 835.31 x (0.5 x 300 - 0.358 x 300) = 35.71 kN-m
Muy1 = Muc + Mus(Total)
= 35.71 + (19.64)
= 55.35 kN-m
Pu/Puz = 0.662
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.662, an = 1.769
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((21.32/76.16)1.769) + ((25.26/55.35)1.769)
= 0.355 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG56 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1534.00 kN
MomentX,(Mx) = 11.33 kN-m
MomentY,(My) = 22.49 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C640.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 31.754 kN.m
My_MinEccen = 30.680 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 31.754 kN.m
My = max(My,My_MinEccen) + MuaddY = 30.680 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 11.33/1534.00 = 7 mm
Actual eccenY = My / P = 22.49/1534.00 = 15 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(7,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(15,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 8 nos. + #16 - 6 nos. (2111 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00307 354.50 8.92 345.58 138.96 177 24.60
2 226 0.00263 348.41 8.92 339.49 76.79 117 9.01
3 226 0.00220 335.19 8.92 326.27 73.80 60 4.40
4 402 0.00176 315.00 8.79 306.20 123.13 -0 -0.00
5 226 0.00132 264.59 7.90 256.69 58.06 -60 -3.46
6 226 0.00090 179.54 6.21 173.33 39.21 -117 -4.60
7 402 0.00046 91.54 3.62 87.92 35.35 -177 -6.26
Total545.31 23.69
xux = 464 mm               Puc = C1.fck.B.D
ku = 1.031
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.027
C1 = 0.446 x (1 - C3/6) = 0.370
Puc = 0.370 x 20.00 x 300 x 450
= 998.00 kN
Pux1 = Puc + Pus(Total)
= 998.00 + (545.31)
= 1543.31 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.426
Muc = 998.00 x (0.5 x 450 - 0.426 x 450) = 33.14 kN-m
Mux1 = Muc + Mus(Total)
= 33.14 + (23.69)
= 56.83 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1056 0.00283 352.39 8.92 343.47 362.55 102 36.98
2 1056 0.00073 146.64 5.34 141.30 149.15 -102 -15.21
Total511.70 21.77
xuy = 323 mm               Puc = C1.fck.B.D
ku = 1.078
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.884
C1 = 0.446 x (1 - C3/6) = 0.380
Puc = 0.380 x 20.00 x 450 x 300
= 1026.69 kN
Puy1 = Puc + Pus(Total)
= 1026.69 + (511.70)
= 1538.39 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.438
Muc = 1026.69 x (0.5 x 300 - 0.438 x 300) = 19.02 kN-m
Muy1 = Muc + Mus(Total)
= 19.02 + (21.77)
= 40.79 kN-m
Pu/Puz = 0.828
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.828, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((31.75/56.83)2.000) + ((30.68/40.79)2.000)
= 0.878 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG56 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 8 nos. + #16 - 6 nos. (2111 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 2019.08 kN
MomentX,(Mx) = 11.04 kN-m
MomentY,(My) = 18.16 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C640.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 41.795 kN.m
My_MinEccen = 40.382 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 41.795 kN.m
My = max(My,My_MinEccen) + MuaddY = 40.382 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 11.04/2019.08 = 5 mm
Actual eccenY = My / P = 18.16/2019.08 = 9 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(5,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(9,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 14 nos. (4398 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00310 354.73 8.92 345.81 217.28 175 38.02
2 628 0.00262 348.28 8.92 339.36 213.23 117 24.88
3 628 0.00215 333.29 8.92 324.37 203.81 58 11.89
4 628 0.00168 309.75 8.69 301.06 189.16 0 0.00
5 628 0.00121 241.34 7.52 233.83 146.92 -58 -8.57
6 628 0.00073 146.91 5.35 141.56 88.95 -117 -10.38
7 628 0.00026 52.48 2.19 50.30 31.60 -175 -5.53
Total1090.94 50.31
xux = 432 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 432/450 = 0.346
Puc = 0.346 x 20.00 x 300 x 450
= 934.03 kN
Pux1 = Puc + Pus(Total)
= 934.03 + (1090.94)
= 2024.97 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 432/450
= 0.400
Muc = 934.03 x (0.5 x 450 - 0.400 x 450) = 42.14 kN-m
Mux1 = Muc + Mus(Total)
= 42.14 + (50.31)
= 92.45 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 2199 0.00287 352.75 8.92 343.83 756.13 100 75.61
2 2199 0.00066 131.36 4.90 126.46 278.10 -100 -27.81
Total1034.24 47.80
xuy = 309 mm               Puc = C1.fck.B.D
ku = 1.031
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.027
C1 = 0.446 x (1 - C3/6) = 0.370
Puc = 0.370 x 20.00 x 450 x 300
= 998.00 kN
Puy1 = Puc + Pus(Total)
= 998.00 + (1034.24)
= 2032.23 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.426
Muc = 998.00 x (0.5 x 300 - 0.426 x 300) = 22.09 kN-m
Muy1 = Muc + Mus(Total)
= 22.09 + (47.80)
= 69.90 kN-m
Pu/Puz = 0.794
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.794, an = 1.989
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((41.79/92.45)1.989) + ((40.38/69.90)1.989)
= 0.542 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 20 = 320 mm
               iii) 300 mm
               Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG56 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 14 nos. (4398 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 2474.40 kN
MomentX,(Mx) = 7.25 kN-m
MomentY,(My) = 9.82 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C640.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 53.818 kN.m
My_MinEccen = 49.488 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 53.818 kN.m
My = max(My,My_MinEccen) + MuaddY = 49.488 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 7.25/2474.40 = 3 mm
Actual eccenY = My / P = 9.82/2474.40 = 4 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(3,22) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(4,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #25 - 12 nos. (5890 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 982 0.00309 354.70 8.92 345.78 339.47 173 58.56
2 982 0.00256 346.53 8.92 337.61 331.45 104 34.30
3 982 0.00202 328.41 8.92 319.49 313.66 35 10.82
4 982 0.00148 292.65 8.32 284.33 279.14 -35 -9.63
5 982 0.00095 189.00 6.44 182.56 179.23 -104 -18.55
6 982 0.00041 81.67 3.27 78.40 76.97 -173 -13.28
Total1519.91 62.23
xux = 450 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 450/450 = 0.360
Puc = 0.360 x 20.00 x 300 x 450
= 972.00 kN
Pux1 = Puc + Pus(Total)
= 972.00 + (1519.91)
= 2491.91 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 450/450
= 0.416
Muc = 972.00 x (0.5 x 450 - 0.416 x 450) = 36.74 kN-m
Mux1 = Muc + Mus(Total)
= 36.74 + (62.23)
= 98.97 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 2945 0.00278 351.98 8.92 343.06 1010.40 98 98.51
2 2945 0.00078 155.88 5.60 150.28 442.61 -98 -43.15
Total1453.01 55.36
xuy = 323 mm               Puc = C1.fck.B.D
ku = 1.078
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.884
C1 = 0.446 x (1 - C3/6) = 0.380
Puc = 0.380 x 20.00 x 450 x 300
= 1026.69 kN
Puy1 = Puc + Pus(Total)
= 1026.69 + (1453.01)
= 2479.69 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.438
Muc = 1026.69 x (0.5 x 300 - 0.438 x 300) = 19.02 kN-m
Muy1 = Muc + Mus(Total)
= 19.02 + (55.36)
= 74.38 kN-m
Pu/Puz = 0.826
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.826, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((53.82/98.97)2.000) + ((49.49/74.38)2.000)
= 0.738 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 25/4 = 6 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (6, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 25 = 400 mm
               iii) 300 mm
               Required spacing = minimum of (300, 400, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG56 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #25 - 12 nos. (5890 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG57 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 342.21 kN
MomentX,(Mx) = 6.41 kN-m
MomentY,(My) = 67.72 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C650.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 7.084 kN.m
My_MinEccen = 6.844 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 7.084 kN.m
My = max(My,My_MinEccen) + MuaddY = 67.718 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 6.41/342.21 = 19 mm
Actual eccenY = My / P = 67.72/342.21 = 198 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(19,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(198,20) = 198 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(198 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00265 349.02 8.92 340.10 76.93 179 13.77
2 226 0.00100 200.38 6.70 193.68 43.81 90 3.92
3 226 -0.00065 -129.63 0.00 -129.63 -29.32 0 0.00
4 226 -0.00230 -338.69 0.00 -338.69 -76.61 -90 6.86
5 226 -0.00395 -360.90 0.00 -360.90 -81.63 -179 14.61
Total-66.83 39.16
xux = 190 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 190/450 = 0.152
Puc = 0.152 x 20.00 x 300 x 450
= 410.06 kN
Pux1 = Puc + Pus(Total)
= 410.06 + (-66.83)
= 343.24 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 190/450
= 0.176
Muc = 410.06 x (0.5 x 450 - 0.176 x 450) = 59.88 kN-m
Mux1 = Muc + Mus(Total)
= 59.88 + (39.16)
= 99.04 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00208 330.54 8.92 321.62 181.87 104 18.91
2 565 -0.00436 -360.90 0.00 -360.90 -204.08 -104 21.22
Total-22.21 40.14
xuy = 113 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 113/300 = 0.136
Puc = 0.136 x 20.00 x 450 x 300
= 366.40 kN
Puy1 = Puc + Pus(Total)
= 366.40 + (-22.21)
= 344.19 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 113/300
= 0.157
Muc = 366.40 x (0.5 x 300 - 0.157 x 300) = 37.72 kN-m
Muy1 = Muc + Mus(Total)
= 37.72 + (40.14)
= 77.86 kN-m
Pu/Puz = 0.220
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.220, an = 1.033
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((7.08/99.04)1.033) + ((67.72/77.86)1.033)
= 0.931 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG57 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 715.84 kN
MomentX,(Mx) = 7.63 kN-m
MomentY,(My) = 62.02 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C650.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 14.818 kN.m
My_MinEccen = 14.317 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 14.818 kN.m
My = max(My,My_MinEccen) + MuaddY = 62.020 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 7.63/715.84 = 11 mm
Actual eccenY = My / P = 62.02/715.84 = 87 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(11,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(87,20) = 87 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(87 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00294 353.36 8.92 344.44 77.91 179 13.95
2 226 0.00184 320.11 8.87 311.24 70.40 90 6.30
3 226 0.00075 150.31 5.44 144.86 32.77 0 0.00
4 226 -0.00034 -68.35 0.00 -68.35 -15.46 -90 1.38
5 226 -0.00144 -287.00 0.00 -287.00 -64.92 -179 11.62
Total100.70 33.25
xux = 287 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 287/450 = 0.229
Puc = 0.229 x 20.00 x 300 x 450
= 618.89 kN
Pux1 = Puc + Pus(Total)
= 618.89 + (100.70)
= 719.59 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 287/450
= 0.265
Muc = 618.89 x (0.5 x 450 - 0.265 x 450) = 65.48 kN-m
Mux1 = Muc + Mus(Total)
= 65.48 + (33.25)
= 98.73 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00269 349.92 8.92 341.00 192.83 104 20.05
2 565 -0.00099 -197.77 0.00 -197.77 -111.83 -104 11.63
Total81.00 31.69
xuy = 198 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 198/300 = 0.238
Puc = 0.238 x 20.00 x 450 x 300
= 641.67 kN
Puy1 = Puc + Pus(Total)
= 641.67 + (81.00)
= 722.67 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 198/300
= 0.275
Muc = 641.67 x (0.5 x 300 - 0.275 x 300) = 43.39 kN-m
Muy1 = Muc + Mus(Total)
= 43.39 + (31.69)
= 75.07 kN-m
Pu/Puz = 0.460
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.460, an = 1.433
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((14.82/98.73)1.433) + ((62.02/75.07)1.433)
= 0.827 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG57 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1070.57 kN
MomentX,(Mx) = 7.20 kN-m
MomentY,(My) = 58.19 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C650.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 22.161 kN.m
My_MinEccen = 21.411 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 22.161 kN.m
My = max(My,My_MinEccen) + MuaddY = 58.189 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 7.20/1070.57 = 7 mm
Actual eccenY = My / P = 58.19/1070.57 = 54 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(7,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(54,20) = 54 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(54 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 6 nos. + #16 - 4 nos. (1483 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00305 354.33 8.92 345.41 138.90 177 24.58
2 226 0.00221 335.40 8.92 326.48 73.85 88 6.46
3 226 0.00139 277.36 8.08 269.28 60.91 0 0.00
4 226 0.00056 113.00 4.33 108.67 24.58 -88 -2.15
5 402 -0.00028 -55.12 0.00 -55.12 -22.16 -177 3.92
Total276.07 32.82
xux = 373 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 373/450 = 0.298
Puc = 0.298 x 20.00 x 300 x 450
= 804.94 kN
Pux1 = Puc + Pus(Total)
= 804.94 + (276.07)
= 1081.01 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 373/450
= 0.345
Muc = 804.94 x (0.5 x 450 - 0.345 x 450) = 56.33 kN-m
Mux1 = Muc + Mus(Total)
= 56.33 + (32.82)
= 89.14 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 741 0.00284 352.47 8.92 343.55 254.71 102 25.98
2 741 0.00002 3.11 0.14 2.97 2.20 -102 -0.22
Total256.92 25.76
xuy = 253 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 253/300 = 0.304
Puc = 0.304 x 20.00 x 450 x 300
= 820.13 kN
Puy1 = Puc + Pus(Total)
= 820.13 + (256.92)
= 1077.04 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 253/300
= 0.351
Muc = 820.13 x (0.5 x 300 - 0.351 x 300) = 36.66 kN-m
Muy1 = Muc + Mus(Total)
= 36.66 + (25.76)
= 62.42 kN-m
Pu/Puz = 0.644
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.644, an = 1.739
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((22.16/89.14)1.739) + ((58.19/62.42)1.739)
= 0.974 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG57 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 6 nos. + #16 - 4 nos. (1483 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1397.59 kN
MomentX,(Mx) = 6.57 kN-m
MomentY,(My) = 52.45 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C650.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 28.930 kN.m
My_MinEccen = 27.952 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 28.930 kN.m
My = max(My,My_MinEccen) + MuaddY = 52.446 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 6.57/1397.59 = 5 mm
Actual eccenY = My / P = 52.45/1397.59 = 38 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(5,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(38,20) = 38 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(38 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 2 nos. + #16 - 10 nos. (2237 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00310 354.73 8.92 345.81 139.06 177 24.61
2 402 0.00249 344.88 8.92 335.96 135.10 105 14.24
3 226 0.00190 323.78 8.90 314.89 71.23 36 2.55
4 402 0.00132 263.31 7.88 255.43 102.71 -34 -3.47
5 402 0.00071 142.49 5.22 137.27 55.20 -105 -5.82
6 402 0.00011 21.67 0.94 20.73 8.34 -177 -1.48
Total511.63 30.64
xux = 415 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 415/450 = 0.332
Puc = 0.332 x 20.00 x 300 x 450
= 896.06 kN
Pux1 = Puc + Pus(Total)
= 896.06 + (511.63)
= 1407.69 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 415/450
= 0.384
Muc = 896.06 x (0.5 x 450 - 0.384 x 450) = 46.98 kN-m
Mux1 = Muc + Mus(Total)
= 46.98 + (30.64)
= 77.61 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1118 0.00291 353.13 8.92 344.21 384.97 102 39.27
2 1118 0.00042 83.08 3.32 79.76 89.21 -102 -9.10
Total474.18 30.17
xuy = 286 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 286/300 = 0.343
Puc = 0.343 x 20.00 x 450 x 300
= 926.44 kN
Puy1 = Puc + Pus(Total)
= 926.44 + (474.18)
= 1400.61 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 286/300
= 0.397
Muc = 926.44 x (0.5 x 300 - 0.397 x 300) = 28.77 kN-m
Muy1 = Muc + Mus(Total)
= 28.77 + (30.17)
= 58.93 kN-m
Pu/Puz = 0.739
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.739, an = 1.898
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((28.93/77.61)1.898) + ((52.45/58.93)1.898)
= 0.955 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG57 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 2 nos. + #16 - 10 nos. (2237 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 1669.92 kN
MomentX,(Mx) = 5.75 kN-m
MomentY,(My) = 32.09 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C650.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 36.321 kN.m
My_MinEccen = 33.398 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 36.321 kN.m
My = max(My,My_MinEccen) + MuaddY = 33.398 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 5.75/1669.92 = 3 mm
Actual eccenY = My / P = 32.09/1669.92 = 19 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(3,22) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(19,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 14 nos. (2815 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00312 354.96 8.92 346.04 139.15 177 24.63
2 402 0.00265 349.09 8.92 340.17 136.79 118 16.14
3 402 0.00219 334.66 8.92 325.74 130.99 59 7.73
4 402 0.00172 312.46 8.75 303.71 122.13 0 0.00
5 402 0.00126 251.21 7.69 243.52 97.93 -59 -5.78
6 402 0.00079 157.98 5.65 152.32 61.25 -118 -7.23
7 402 0.00032 64.74 2.65 62.09 24.97 -177 -4.42
Total713.20 31.07
xux = 443 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 443/450 = 0.354
Puc = 0.354 x 20.00 x 300 x 450
= 956.81 kN
Pux1 = Puc + Pus(Total)
= 956.81 + (713.20)
= 1670.01 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 443/450
= 0.410
Muc = 956.81 x (0.5 x 450 - 0.410 x 450) = 38.97 kN-m
Mux1 = Muc + Mus(Total)
= 38.97 + (31.07)
= 70.04 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1407 0.00287 352.75 8.92 343.83 483.92 102 49.36
2 1407 0.00067 133.83 4.97 128.86 181.37 -102 -18.50
Total665.29 30.86
xuy = 314 mm               Puc = C1.fck.B.D
ku = 1.047
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.976
C1 = 0.446 x (1 - C3/6) = 0.373
Puc = 0.373 x 20.00 x 450 x 300
= 1008.29 kN
Puy1 = Puc + Pus(Total)
= 1008.29 + (665.29)
= 1673.58 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.431
Muc = 1008.29 x (0.5 x 300 - 0.431 x 300) = 20.99 kN-m
Muy1 = Muc + Mus(Total)
= 20.99 + (30.86)
= 51.85 kN-m
Pu/Puz = 0.808
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.808, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((36.32/70.04)2.000) + ((33.40/51.85)2.000)
= 0.684 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG57 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 14 nos. (2815 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG58 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 435.13 kN
MomentX,(Mx) = 17.61 kN-m
MomentY,(My) = 112.04 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C660.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 9.007 kN.m
My_MinEccen = 8.703 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 17.607 kN.m
My = max(My,My_MinEccen) + MuaddY = 112.041 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 17.61/435.13 = 40 mm
Actual eccenY = My / P = 112.04/435.13 = 257 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(40,21) = 40 mm
eccenY = max(Actual eccenY,eccenYMin) = max(257,20) = 257 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(40 mm) > 0.05 x 300(15 mm)
and eccenY(257 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 14 nos. (2815 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00273 351.13 8.92 342.21 137.61 177 24.36
2 402 0.00179 316.82 8.82 308.00 123.85 118 14.61
3 402 0.00085 170.10 5.97 164.13 66.00 59 3.89
4 402 -0.00009 -18.24 0.00 -18.24 -7.33 0 0.00
5 402 -0.00103 -206.57 0.00 -206.57 -83.07 -59 4.90
6 402 -0.00197 -326.80 0.00 -326.80 -131.42 -118 15.51
7 402 -0.00292 -353.17 0.00 -353.17 -142.02 -177 25.14
Total-36.37 88.41
xux = 219 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 219/450 = 0.175
Puc = 0.175 x 20.00 x 300 x 450
= 473.66 kN
Pux1 = Puc + Pus(Total)
= 473.66 + (-36.37)
= 437.29 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 219/450
= 0.203
Muc = 473.66 x (0.5 x 450 - 0.203 x 450) = 63.36 kN-m
Mux1 = Muc + Mus(Total)
= 63.36 + (88.41)
= 151.78 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1407 0.00232 339.67 8.92 330.75 465.52 102 47.48
2 1407 -0.00267 -349.46 0.00 -349.46 -491.85 -102 50.17
Total-26.33 97.65
xuy = 143 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 143/300 = 0.172
Puc = 0.172 x 20.00 x 450 x 300
= 463.22 kN
Puy1 = Puc + Pus(Total)
= 463.22 + (-26.33)
= 436.89 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 143/300
= 0.198
Muc = 463.22 x (0.5 x 300 - 0.198 x 300) = 41.93 kN-m
Muy1 = Muc + Mus(Total)
= 41.93 + (97.65)
= 139.58 kN-m
Pu/Puz = 0.211
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.211, an = 1.018
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((17.61/151.78)1.018) + ((112.04/139.58)1.018)
= 0.911 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG58 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 14 nos. (2815 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 879.03 kN
MomentX,(Mx) = 14.76 kN-m
MomentY,(My) = 102.30 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C660.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 18.196 kN.m
My_MinEccen = 17.581 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 18.196 kN.m
My = max(My,My_MinEccen) + MuaddY = 102.301 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 14.76/879.03 = 17 mm
Actual eccenY = My / P = 102.30/879.03 = 116 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(17,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(116,20) = 116 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(116 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 14 nos. (2815 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00291 353.11 8.92 344.19 138.41 177 24.50
2 402 0.00218 334.53 8.92 325.61 130.94 118 15.45
3 402 0.00146 290.57 8.27 282.30 113.52 59 6.70
4 402 0.00073 146.91 5.35 141.56 56.93 0 0.00
5 402 0.00001 1.88 0.08 1.80 0.72 -59 -0.04
6 402 -0.00072 -143.15 0.00 -143.15 -57.56 -118 6.79
7 402 -0.00144 -288.79 0.00 -288.79 -116.13 -177 20.55
Total266.82 73.95
xux = 285 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 285/450 = 0.228
Puc = 0.228 x 20.00 x 300 x 450
= 615.09 kN
Pux1 = Puc + Pus(Total)
= 615.09 + (266.82)
= 881.92 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 285/450
= 0.263
Muc = 615.09 x (0.5 x 450 - 0.263 x 450) = 65.53 kN-m
Mux1 = Muc + Mus(Total)
= 65.53 + (73.95)
= 139.48 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1407 0.00267 349.40 8.92 340.48 479.20 102 48.88
2 1407 -0.00088 -175.16 0.00 -175.16 -246.53 -102 25.15
Total232.67 74.02
xuy = 202 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 202/300 = 0.242
Puc = 0.242 x 20.00 x 450 x 300
= 653.06 kN
Puy1 = Puc + Pus(Total)
= 653.06 + (232.67)
= 885.73 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 202/300
= 0.280
Muc = 653.06 x (0.5 x 300 - 0.280 x 300) = 43.20 kN-m
Muy1 = Muc + Mus(Total)
= 43.20 + (74.02)
= 117.22 kN-m
Pu/Puz = 0.426
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.426, an = 1.376
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((18.20/139.48)1.376) + ((102.30/117.22)1.376)
= 0.890 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG58 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 14 nos. (2815 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1332.27 kN
MomentX,(Mx) = 15.56 kN-m
MomentY,(My) = 98.87 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C660.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 27.578 kN.m
My_MinEccen = 26.645 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 27.578 kN.m
My = max(My,My_MinEccen) + MuaddY = 98.866 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 15.56/1332.27 = 12 mm
Actual eccenY = My / P = 98.87/1332.27 = 74 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(12,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(74,20) = 74 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(74 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 12 nos. (3770 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00298 353.69 8.92 344.77 216.63 175 37.91
2 628 0.00224 336.65 8.92 327.73 205.92 105 21.62
3 628 0.00151 295.23 8.38 286.84 180.23 35 6.31
4 628 0.00078 155.06 5.58 149.49 93.93 -35 -3.29
5 628 0.00004 8.35 0.37 7.98 5.02 -105 -0.53
6 628 -0.00069 -138.36 0.00 -138.36 -86.94 -175 15.21
Total614.78 77.24
xux = 334 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 334/450 = 0.267
Puc = 0.267 x 20.00 x 300 x 450
= 721.41 kN
Pux1 = Puc + Pus(Total)
= 721.41 + (614.78)
= 1336.18 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 334/450
= 0.309
Muc = 721.41 x (0.5 x 450 - 0.309 x 450) = 62.09 kN-m
Mux1 = Muc + Mus(Total)
= 62.09 + (77.24)
= 139.32 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1885 0.00276 351.81 8.92 342.89 646.33 100 64.63
2 1885 -0.00020 -39.27 0.00 -39.27 -74.03 -100 7.40
Total572.30 72.04
xuy = 237 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 237/300 = 0.284
Puc = 0.284 x 20.00 x 450 x 300
= 766.97 kN
Puy1 = Puc + Pus(Total)
= 766.97 + (572.30)
= 1339.26 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 237/300
= 0.328
Muc = 766.97 x (0.5 x 300 - 0.328 x 300) = 39.52 kN-m
Muy1 = Muc + Mus(Total)
= 39.52 + (72.04)
= 111.55 kN-m
Pu/Puz = 0.566
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.566, an = 1.610
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((27.58/139.32)1.610) + ((98.87/111.55)1.610)
= 0.897 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 20 = 320 mm
               iii) 300 mm
               Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG58 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 12 nos. (3770 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1801.15 kN
MomentX,(Mx) = 16.28 kN-m
MomentY,(My) = 93.96 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C660.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 37.284 kN.m
My_MinEccen = 36.023 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 37.284 kN.m
My = max(My,My_MinEccen) + MuaddY = 93.959 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 16.28/1801.15 = 9 mm
Actual eccenY = My / P = 93.96/1801.15 = 52 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(9,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(52,20) = 52 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(52 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 16 nos. (5027 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00303 354.13 8.92 345.21 216.90 175 37.96
2 628 0.00255 346.45 8.92 337.53 212.07 125 26.51
3 628 0.00208 330.60 8.92 321.68 202.11 75 15.16
4 628 0.00160 304.21 8.57 295.64 185.76 25 4.64
5 628 0.00113 225.93 7.23 218.70 137.41 -25 -3.44
6 628 0.00066 131.11 4.89 126.22 79.31 -75 -5.95
7 628 0.00018 36.30 1.55 34.75 21.83 -125 -2.73
8 628 -0.00029 -58.52 0.00 -58.52 -36.77 -175 6.43
Total1018.63 78.59
xux = 369 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 369/450 = 0.295
Puc = 0.295 x 20.00 x 300 x 450
= 797.34 kN
Pux1 = Puc + Pus(Total)
= 797.34 + (1018.63)
= 1815.97 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 369/450
= 0.341
Muc = 797.34 x (0.5 x 450 - 0.341 x 450) = 56.96 kN-m
Mux1 = Muc + Mus(Total)
= 56.96 + (78.59)
= 135.55 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 2513 0.00284 352.49 8.92 343.57 863.49 100 86.35
2 2513 0.00020 39.23 1.66 37.57 94.42 -100 -9.44
Total957.92 76.91
xuy = 265 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 265/300 = 0.318
Puc = 0.318 x 20.00 x 450 x 300
= 858.09 kN
Puy1 = Puc + Pus(Total)
= 858.09 + (957.92)
= 1816.01 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 265/300
= 0.367
Muc = 858.09 x (0.5 x 300 - 0.367 x 300) = 34.17 kN-m
Muy1 = Muc + Mus(Total)
= 34.17 + (76.91)
= 111.08 kN-m
Pu/Puz = 0.659
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.659, an = 1.765
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((37.28/135.55)1.765) + ((93.96/111.08)1.765)
= 0.847 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 20 = 320 mm
               iii) 300 mm
               Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG58 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 16 nos. (5027 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 2301.10 kN
MomentX,(Mx) = 10.77 kN-m
MomentY,(My) = 62.49 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C660.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 50.049 kN.m
My_MinEccen = 46.022 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 50.049 kN.m
My = max(My,My_MinEccen) + MuaddY = 62.490 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 10.77/2301.10 = 5 mm
Actual eccenY = My / P = 62.49/2301.10 = 27 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(5,22) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(27,20) = 27 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(27 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #25 - 12 nos. (5890 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 982 0.00306 354.43 8.92 345.51 339.21 173 58.51
2 982 0.00248 344.69 8.92 335.77 329.64 104 34.12
3 982 0.00191 323.94 8.90 315.04 309.29 35 10.67
4 982 0.00133 265.80 7.92 257.89 253.18 -35 -8.73
5 982 0.00075 150.35 5.45 144.91 142.26 -104 -14.72
6 982 0.00017 34.90 1.49 33.41 32.80 -173 -5.66
Total1406.39 74.18
xux = 418 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 418/450 = 0.335
Puc = 0.335 x 20.00 x 300 x 450
= 903.66 kN
Pux1 = Puc + Pus(Total)
= 903.66 + (1406.39)
= 2310.04 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 418/450
= 0.387
Muc = 903.66 x (0.5 x 450 - 0.387 x 450) = 46.05 kN-m
Mux1 = Muc + Mus(Total)
= 46.05 + (74.18)
= 120.24 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 2945 0.00288 352.87 8.92 343.95 1013.03 98 98.77
2 2945 0.00059 117.95 4.49 113.47 334.19 -98 -32.58
Total1347.22 66.19
xuy = 298 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 298/300 = 0.357
Puc = 0.357 x 20.00 x 450 x 300
= 964.41 kN
Puy1 = Puc + Pus(Total)
= 964.41 + (1347.22)
= 2311.62 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 298/300
= 0.413
Muc = 964.41 x (0.5 x 300 - 0.413 x 300) = 25.24 kN-m
Muy1 = Muc + Mus(Total)
= 25.24 + (66.19)
= 91.43 kN-m
Pu/Puz = 0.768
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.768, an = 1.947
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((50.05/120.24)1.947) + ((62.49/91.43)1.947)
= 0.658 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 25/4 = 6 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (6, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 25 = 400 mm
               iii) 300 mm
               Required spacing = minimum of (300, 400, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG58 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #25 - 12 nos. (5890 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG59 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 748.64 kN
MomentX,(Mx) = 123.80 kN-m
MomentY,(My) = 33.63 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C670.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 15.497 kN.m
My_MinEccen = 14.973 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 123.804 kN.m
My = max(My,My_MinEccen) + MuaddY = 33.631 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 123.80/748.64 = 165 mm
Actual eccenY = My / P = 33.63/748.64 = 45 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(165,21) = 165 mm
eccenY = max(Actual eccenY,eccenYMin) = max(45,20) = 45 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(165 mm) > 0.05 x 300(15 mm)
and eccenY(45 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 16 nos. (3217 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00286 352.64 8.92 343.72 138.22 177 24.46
2 402 0.00218 334.29 8.92 325.37 130.84 126 16.54
3 402 0.00150 294.41 8.36 286.05 115.03 76 8.73
4 402 0.00082 164.44 5.83 158.61 63.78 25 1.61
5 402 0.00014 28.83 1.24 27.59 11.09 -25 -0.28
6 402 -0.00053 -106.79 0.00 -106.79 -42.94 -76 3.26
7 402 -0.00121 -242.40 0.00 -242.40 -97.48 -126 12.32
8 402 -0.00189 -322.93 0.00 -322.93 -129.86 -177 22.99
Total188.68 89.63
xux = 261 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 261/450 = 0.209
Puc = 0.209 x 20.00 x 300 x 450
= 563.84 kN
Pux1 = Puc + Pus(Total)
= 563.84 + (188.68)
= 752.52 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 261/450
= 0.241
Muc = 563.84 x (0.5 x 450 - 0.241 x 450) = 65.64 kN-m
Mux1 = Muc + Mus(Total)
= 65.64 + (89.63)
= 155.27 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1608 0.00260 347.64 8.92 338.72 544.84 102 55.57
2 1608 -0.00123 -246.72 0.00 -246.72 -396.84 -102 40.48
Total147.99 96.05
xuy = 186 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 186/300 = 0.224
Puc = 0.224 x 20.00 x 450 x 300
= 603.70 kN
Puy1 = Puc + Pus(Total)
= 603.70 + (147.99)
= 751.70 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 186/300
= 0.258
Muc = 603.70 x (0.5 x 300 - 0.258 x 300) = 43.76 kN-m
Muy1 = Muc + Mus(Total)
= 43.76 + (96.05)
= 139.81 kN-m
Pu/Puz = 0.342
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.342, an = 1.237
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((123.80/155.27)1.237) + ((33.63/139.81)1.237)
= 0.927 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG59 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 16 nos. (3217 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1461.55 kN
MomentX,(Mx) = 109.55 kN-m
MomentY,(My) = 30.32 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C670.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 30.254 kN.m
My_MinEccen = 29.231 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 109.550 kN.m
My = max(My,My_MinEccen) + MuaddY = 30.324 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 109.55/1461.55 = 75 mm
Actual eccenY = My / P = 30.32/1461.55 = 21 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(75,21) = 75 mm
eccenY = max(Actual eccenY,eccenYMin) = max(21,20) = 21 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(75 mm) > 0.05 x 300(15 mm)
and eccenY(21 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 12 nos. (3770 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00301 353.96 8.92 345.04 216.80 175 37.94
2 628 0.00232 339.39 8.92 330.47 207.64 105 21.80
3 628 0.00163 306.43 8.61 297.82 187.13 35 6.55
4 628 0.00094 187.44 6.40 181.04 113.75 -35 -3.98
5 628 0.00025 49.44 2.07 47.37 29.76 -105 -3.13
6 628 -0.00044 -88.56 0.00 -88.56 -55.64 -175 9.74
Total699.43 68.92
xux = 355 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 355/450 = 0.284
Puc = 0.284 x 20.00 x 300 x 450
= 766.97 kN
Pux1 = Puc + Pus(Total)
= 766.97 + (699.43)
= 1466.40 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 355/450
= 0.328
Muc = 766.97 x (0.5 x 450 - 0.328 x 450) = 59.28 kN-m
Mux1 = Muc + Mus(Total)
= 59.28 + (68.92)
= 128.20 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1885 0.00280 352.17 8.92 343.25 647.01 100 64.70
2 1885 0.00001 2.18 0.10 2.08 3.93 -100 -0.39
Total650.94 64.31
xuy = 251 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 251/300 = 0.301
Puc = 0.301 x 20.00 x 450 x 300
= 812.53 kN
Puy1 = Puc + Pus(Total)
= 812.53 + (650.94)
= 1463.47 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 251/300
= 0.348
Muc = 812.53 x (0.5 x 300 - 0.348 x 300) = 37.11 kN-m
Muy1 = Muc + Mus(Total)
= 37.11 + (64.31)
= 101.42 kN-m
Pu/Puz = 0.621
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.621, an = 1.701
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((109.55/128.20)1.701) + ((30.32/101.42)1.701)
= 0.894 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 20 = 320 mm
               iii) 300 mm
               Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG59 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 12 nos. (3770 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 2227.39 kN
MomentX,(Mx) = 101.19 kN-m
MomentY,(My) = 30.36 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C670.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 46.107 kN.m
My_MinEccen = 44.548 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 101.194 kN.m
My = max(My,My_MinEccen) + MuaddY = 44.548 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 101.19/2227.39 = 45 mm
Actual eccenY = My / P = 30.36/2227.39 = 14 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(45,21) = 45 mm
eccenY = max(Actual eccenY,eccenYMin) = max(14,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(45 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #25 - 12 nos. (5890 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 982 0.00305 354.33 8.92 345.41 339.11 173 58.50
2 982 0.00246 344.01 8.92 335.09 328.98 104 34.05
3 982 0.00187 321.37 8.88 312.49 306.79 35 10.58
4 982 0.00127 254.57 7.74 246.83 242.33 -35 -8.36
5 982 0.00068 136.14 5.04 131.10 128.71 -104 -13.32
6 982 0.00009 17.70 0.77 16.93 16.62 -173 -2.87
Total1362.53 78.58
xux = 408 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 408/450 = 0.326
Puc = 0.326 x 20.00 x 300 x 450
= 880.88 kN
Pux1 = Puc + Pus(Total)
= 880.88 + (1362.53)
= 2243.40 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 408/450
= 0.377
Muc = 880.88 x (0.5 x 450 - 0.377 x 450) = 48.76 kN-m
Mux1 = Muc + Mus(Total)
= 48.76 + (78.58)
= 127.34 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 2945 0.00287 352.74 8.92 343.82 1012.64 98 98.73
2 2945 0.00052 103.87 4.03 99.84 294.05 -98 -28.67
Total1306.69 70.06
xuy = 291 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 291/300 = 0.349
Puc = 0.349 x 20.00 x 450 x 300
= 941.63 kN
Puy1 = Puc + Pus(Total)
= 941.63 + (1306.69)
= 2248.32 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 291/300
= 0.403
Muc = 941.63 x (0.5 x 300 - 0.403 x 300) = 27.40 kN-m
Muy1 = Muc + Mus(Total)
= 27.40 + (70.06)
= 97.46 kN-m
Pu/Puz = 0.744
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.744, an = 1.906
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((101.19/127.34)1.906) + ((44.55/97.46)1.906)
= 0.870 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 25/4 = 6 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (6, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 25 = 400 mm
               iii) 300 mm
               Required spacing = minimum of (300, 400, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG59 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #25 - 12 nos. (5890 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 3074.59 kN
MomentX,(Mx) = 87.11 kN-m
MomentY,(My) = 30.63 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C670.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 63.644 kN.m
My_MinEccen = 61.492 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 87.107 kN.m
My = max(My,My_MinEccen) + MuaddY = 61.492 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 87.11/3074.59 = 28 mm
Actual eccenY = My / P = 30.63/3074.59 = 10 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(28,21) = 28 mm
eccenY = max(Actual eccenY,eccenYMin) = max(10,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(28 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #32 - 12 nos. (9651 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1608 0.00302 354.03 8.92 345.11 555.11 169 93.81
2 1608 0.00243 343.31 8.92 334.39 537.87 101 54.54
3 1608 0.00184 320.11 8.87 311.24 500.63 34 16.92
4 1608 0.00126 251.91 7.70 244.22 392.82 -34 -13.28
5 1608 0.00067 134.87 5.00 129.87 208.89 -101 -21.18
6 1608 0.00009 17.83 0.78 17.05 27.43 -169 -4.63
Total2222.75 126.18
xux = 404 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 404/450 = 0.323
Puc = 0.323 x 20.00 x 300 x 450
= 873.28 kN
Pux1 = Puc + Pus(Total)
= 873.28 + (2222.75)
= 3096.03 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 404/450
= 0.374
Muc = 873.28 x (0.5 x 450 - 0.374 x 450) = 49.61 kN-m
Mux1 = Muc + Mus(Total)
= 49.61 + (126.18)
= 175.79 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 4825 0.00282 352.33 8.92 343.41 1657.10 94 155.77
2 4825 0.00054 107.52 4.15 103.37 498.82 -94 -46.89
Total2155.92 108.88
xuy = 288 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 288/300 = 0.346
Puc = 0.346 x 20.00 x 450 x 300
= 934.03 kN
Puy1 = Puc + Pus(Total)
= 934.03 + (2155.92)
= 3089.95 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 288/300
= 0.400
Muc = 934.03 x (0.5 x 300 - 0.400 x 300) = 28.09 kN-m
Muy1 = Muc + Mus(Total)
= 28.09 + (108.88)
= 136.97 kN-m
Pu/Puz = 0.744
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.744, an = 1.907
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((87.11/175.79)1.907) + ((61.49/136.97)1.907)
= 0.479 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 32/4 = 8 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (8, 6) = 8 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 32 = 512 mm
               iii) 300 mm
               Required spacing = minimum of (300, 512, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG59 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #32 - 12 nos. (9651 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 600 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 4064.54 kN
MomentX,(Mx) = 49.26 kN-m
MomentY,(My) = 19.80 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C670.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 600 = 5.63 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 108.727 kN.m
My_MinEccen = 81.291 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 108.727 kN.m
My = max(My,My_MinEccen) + MuaddY = 81.291 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 49.26/4064.54 = 12 mm
Actual eccenY = My / P = 19.80/4064.54 = 5 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (600/30) = 27 mm
eccenXMin = max(27,20)= 27 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 27 mm
eccenX = max(Actual eccenX,eccenXMin) = max(12,27) = 27 mm
eccenY = max(Actual eccenY,eccenYMin) = max(5,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(27 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 600(30 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #32 - 14 nos. (11259 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1608 0.00317 355.37 8.92 346.45 557.27 244 135.97
2 1608 0.00269 349.90 8.92 340.98 548.47 163 89.22
3 1608 0.00220 335.24 8.92 326.32 524.88 81 42.69
4 1608 0.00172 312.46 8.75 303.71 488.51 0 0.00
5 1608 0.00124 248.05 7.63 240.42 386.71 -81 -31.45
6 1608 0.00076 151.65 5.48 146.17 235.12 -163 -38.25
7 1608 0.00028 55.26 2.29 52.96 85.19 -244 -20.79
Total2826.15 177.40
xux = 591 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 591/600 = 0.354
Puc = 0.354 x 20.00 x 300 x 600
= 1275.75 kN
Pux1 = Puc + Pus(Total)
= 1275.75 + (2826.15)
= 4101.90 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 591/600
= 0.410
Muc = 1275.75 x (0.5 x 600 - 0.410 x 600) = 69.27 kN-m
Mux1 = Muc + Mus(Total)
= 69.27 + (177.40)
= 246.67 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 5630 0.00278 352.00 8.92 343.08 1931.43 94 181.55
2 5630 0.00076 151.09 5.47 145.62 819.80 -94 -77.06
Total2751.23 104.49
xuy = 314 mm               Puc = C1.fck.B.D
ku = 1.047
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.976
C1 = 0.446 x (1 - C3/6) = 0.373
Puc = 0.373 x 20.00 x 600 x 300
= 1344.38 kN
Puy1 = Puc + Pus(Total)
= 1344.38 + (2751.23)
= 4095.62 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.431
Muc = 1344.38 x (0.5 x 300 - 0.431 x 300) = 27.99 kN-m
Muy1 = Muc + Mus(Total)
= 27.99 + (104.49)
= 132.48 kN-m
Pu/Puz = 0.809
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.809, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((108.73/246.67)2.000) + ((81.29/132.48)2.000)
= 0.571 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 32/4 = 8 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (8, 6) = 8 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 32 = 512 mm
               iii) 300 mm
               Required spacing = minimum of (300, 512, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG59 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 600 mm
Provide #32 - 14 nos. (11259 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG60 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 640.09 kN
MomentX,(Mx) = 102.01 kN-m
MomentY,(My) = 1.72 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C680.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 13.250 kN.m
My_MinEccen = 12.802 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 102.007 kN.m
My = max(My,My_MinEccen) + MuaddY = 12.802 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 102.01/640.09 = 159 mm
Actual eccenY = My / P = 1.72/640.09 = 3 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(159,21) = 159 mm
eccenY = max(Actual eccenY,eccenYMin) = max(3,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(159 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 6 nos. (1206 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00288 352.84 8.92 343.92 138.30 177 24.48
2 402 0.00059 118.18 4.49 113.69 45.72 0 0.00
3 402 -0.00170 -310.92 0.00 -310.92 -125.03 -177 22.13
Total58.99 46.61
xux = 271 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 271/450 = 0.217
Puc = 0.217 x 20.00 x 300 x 450
= 584.72 kN
Pux1 = Puc + Pus(Total)
= 584.72 + (58.99)
= 643.71 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 271/450
= 0.250
Muc = 584.72 x (0.5 x 450 - 0.250 x 450) = 65.72 kN-m
Mux1 = Muc + Mus(Total)
= 65.72 + (46.61)
= 112.32 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 603 0.00259 347.35 8.92 338.43 204.14 102 20.82
2 603 -0.00129 -258.78 0.00 -258.78 -156.09 -102 15.92
Total48.04 36.74
xuy = 184 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 184/300 = 0.221
Puc = 0.221 x 20.00 x 450 x 300
= 596.11 kN
Puy1 = Puc + Pus(Total)
= 596.11 + (48.04)
= 644.15 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 184/300
= 0.255
Muc = 596.11 x (0.5 x 300 - 0.255 x 300) = 43.79 kN-m
Muy1 = Muc + Mus(Total)
= 43.79 + (36.74)
= 80.53 kN-m
Pu/Puz = 0.405
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.405, an = 1.342
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((102.01/112.32)1.342) + ((12.80/80.53)1.342)
= 0.963 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG60 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 6 nos. (1206 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1272.08 kN
MomentX,(Mx) = 89.92 kN-m
MomentY,(My) = 1.19 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C680.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 26.332 kN.m
My_MinEccen = 25.442 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 89.917 kN.m
My = max(My,My_MinEccen) + MuaddY = 25.442 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 89.92/1272.08 = 71 mm
Actual eccenY = My / P = 1.19/1272.08 = 1 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(71,21) = 71 mm
eccenY = max(Actual eccenY,eccenYMin) = max(1,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(71 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 12 nos. (2413 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00304 354.29 8.92 345.37 138.88 177 24.58
2 402 0.00237 341.46 8.92 332.54 133.72 106 14.20
3 402 0.00170 311.21 8.72 302.49 121.64 35 4.31
4 402 0.00103 206.20 6.83 199.38 80.17 -35 -2.84
5 402 0.00036 71.95 2.92 69.03 27.76 -106 -2.95
6 402 -0.00031 -62.31 0.00 -62.31 -25.06 -177 4.44
Total477.12 41.74
xux = 369 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 369/450 = 0.295
Puc = 0.295 x 20.00 x 300 x 450
= 797.34 kN
Pux1 = Puc + Pus(Total)
= 797.34 + (477.12)
= 1274.46 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 369/450
= 0.341
Muc = 797.34 x (0.5 x 450 - 0.341 x 450) = 56.96 kN-m
Mux1 = Muc + Mus(Total)
= 56.96 + (41.74)
= 98.70 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1206 0.00285 352.62 8.92 343.70 414.64 102 42.29
2 1206 0.00011 21.95 0.95 20.99 25.33 -102 -2.58
Total439.96 39.71
xuy = 260 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 260/300 = 0.312
Puc = 0.312 x 20.00 x 450 x 300
= 842.91 kN
Puy1 = Puc + Pus(Total)
= 842.91 + (439.96)
= 1282.87 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 260/300
= 0.361
Muc = 842.91 x (0.5 x 300 - 0.361 x 300) = 35.21 kN-m
Muy1 = Muc + Mus(Total)
= 35.21 + (39.71)
= 74.92 kN-m
Pu/Puz = 0.654
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.654, an = 1.757
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((89.92/98.70)1.757) + ((25.44/74.92)1.757)
= 0.999 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG60 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 12 nos. (2413 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1942.07 kN
MomentX,(Mx) = 82.35 kN-m
MomentY,(My) = 0.72 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C680.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 40.201 kN.m
My_MinEccen = 38.841 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 82.349 kN.m
My = max(My,My_MinEccen) + MuaddY = 38.841 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 82.35/1942.07 = 42 mm
Actual eccenY = My / P = 0.72/1942.07 = 0 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(42,21) = 42 mm
eccenY = max(Actual eccenY,eccenYMin) = max(0,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(42 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 14 nos. (4398 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00308 354.61 8.92 345.69 217.21 175 38.01
2 628 0.00259 347.52 8.92 338.60 212.75 117 24.82
3 628 0.00211 331.62 8.92 322.70 202.76 58 11.83
4 628 0.00162 305.53 8.59 296.94 186.57 0 0.00
5 628 0.00113 225.93 7.23 218.70 137.41 -58 -8.02
6 628 0.00064 128.32 4.81 123.52 77.61 -117 -9.05
7 628 0.00015 30.72 1.32 29.40 18.47 -175 -3.23
Total1052.78 54.36
xux = 418 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 418/450 = 0.335
Puc = 0.335 x 20.00 x 300 x 450
= 903.66 kN
Pux1 = Puc + Pus(Total)
= 903.66 + (1052.78)
= 1956.43 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 418/450
= 0.387
Muc = 903.66 x (0.5 x 450 - 0.387 x 450) = 46.05 kN-m
Mux1 = Muc + Mus(Total)
= 46.05 + (54.36)
= 100.41 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 2199 0.00291 353.13 8.92 344.21 756.96 100 75.70
2 2199 0.00056 112.07 4.30 107.78 237.01 -100 -23.70
Total993.97 51.99
xuy = 298 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 298/300 = 0.357
Puc = 0.357 x 20.00 x 450 x 300
= 964.41 kN
Puy1 = Puc + Pus(Total)
= 964.41 + (993.97)
= 1958.38 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 298/300
= 0.413
Muc = 964.41 x (0.5 x 300 - 0.413 x 300) = 25.24 kN-m
Muy1 = Muc + Mus(Total)
= 25.24 + (51.99)
= 77.24 kN-m
Pu/Puz = 0.763
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.763, an = 1.939
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((82.35/100.41)1.939) + ((38.84/77.24)1.939)
= 0.945 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 20 = 320 mm
               iii) 300 mm
               Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG60 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 14 nos. (4398 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 2673.67 kN
MomentX,(Mx) = 69.22 kN-m
MomentY,(My) = 0.12 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C680.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 55.345 kN.m
My_MinEccen = 53.473 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 69.224 kN.m
My = max(My,My_MinEccen) + MuaddY = 53.473 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 69.22/2673.67 = 26 mm
Actual eccenY = My / P = 0.12/2673.67 = 0 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(26,21) = 26 mm
eccenY = max(Actual eccenY,eccenYMin) = max(0,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(26 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #25 - 14 nos. (6872 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 982 0.00308 354.62 8.92 345.70 339.39 173 58.54
2 982 0.00262 348.30 8.92 339.38 333.19 115 38.32
3 982 0.00217 333.84 8.92 324.92 318.98 58 18.34
4 982 0.00171 311.57 8.73 302.84 297.31 0 0.00
5 982 0.00125 250.01 7.67 242.34 237.92 -58 -13.68
6 982 0.00079 158.42 5.67 152.75 149.96 -115 -17.25
7 982 0.00033 66.83 2.73 64.10 62.93 -173 -10.85
Total1739.68 73.42
xux = 439 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 439/450 = 0.352
Puc = 0.352 x 20.00 x 300 x 450
= 949.22 kN
Pux1 = Puc + Pus(Total)
= 949.22 + (1739.68)
= 2688.90 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 439/450
= 0.406
Muc = 949.22 x (0.5 x 450 - 0.406 x 450) = 40.05 kN-m
Mux1 = Muc + Mus(Total)
= 40.05 + (73.42)
= 113.47 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 3436 0.00280 352.15 8.92 343.23 1179.38 98 114.99
2 3436 0.00075 149.86 5.43 144.43 496.27 -98 -48.39
Total1675.65 66.60
xuy = 319 mm               Puc = C1.fck.B.D
ku = 1.063
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.929
C1 = 0.446 x (1 - C3/6) = 0.377
Puc = 0.377 x 20.00 x 450 x 300
= 1017.83 kN
Puy1 = Puc + Pus(Total)
= 1017.83 + (1675.65)
= 2693.48 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.435
Muc = 1017.83 x (0.5 x 300 - 0.435 x 300) = 19.97 kN-m
Muy1 = Muc + Mus(Total)
= 19.97 + (66.60)
= 86.57 kN-m
Pu/Puz = 0.812
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.812, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((69.22/113.47)2.000) + ((53.47/86.57)2.000)
= 0.754 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 25/4 = 6 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (6, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 25 = 400 mm
               iii) 300 mm
               Required spacing = minimum of (300, 400, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG60 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #25 - 14 nos. (6872 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 3513.33 kN
MomentX,(Mx) = 36.15 kN-m
MomentY,(My) = 0.78 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C680.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 76.415 kN.m
My_MinEccen = 70.267 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 76.415 kN.m
My = max(My,My_MinEccen) + MuaddY = 70.267 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 36.15/3513.33 = 10 mm
Actual eccenY = My / P = 0.78/3513.33 = 0 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(10,22) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(0,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #40 - 8 nos. (10053 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 2513 0.00303 354.19 8.92 345.27 867.76 165 143.18
2 2513 0.00218 334.27 8.92 325.35 817.69 55 44.97
3 2513 0.00132 264.44 7.90 256.55 644.78 -55 -35.46
4 2513 0.00047 93.33 3.68 89.66 225.33 -165 -37.18
Total2555.56 115.51
xux = 450 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 450/450 = 0.360
Puc = 0.360 x 20.00 x 300 x 450
= 972.00 kN
Pux1 = Puc + Pus(Total)
= 972.00 + (2555.56)
= 3527.56 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 450/450
= 0.416
Muc = 972.00 x (0.5 x 450 - 0.416 x 450) = 36.74 kN-m
Mux1 = Muc + Mus(Total)
= 36.74 + (115.51)
= 152.25 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 5027 0.00272 350.80 8.92 341.88 1718.48 90 154.66
2 5027 0.00083 165.63 5.86 159.78 803.12 -90 -72.28
Total2521.60 82.38
xuy = 319 mm               Puc = C1.fck.B.D
ku = 1.063
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.929
C1 = 0.446 x (1 - C3/6) = 0.377
Puc = 0.377 x 20.00 x 450 x 300
= 1017.83 kN
Puy1 = Puc + Pus(Total)
= 1017.83 + (2521.60)
= 3539.43 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.435
Muc = 1017.83 x (0.5 x 300 - 0.435 x 300) = 19.97 kN-m
Muy1 = Muc + Mus(Total)
= 19.97 + (82.38)
= 102.35 kN-m
Pu/Puz = 0.826
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.826, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((76.41/152.25)2.000) + ((70.27/102.35)2.000)
= 0.723 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 40/4 = 10 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (10, 6) = 10 mm
               Provide 10 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 40 = 640 mm
               iii) 300 mm
               Required spacing = minimum of (300, 640, 300) = 300 mm
# Provide ties               10 @ 300 mm c/c

SUMMARY :

CG60 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #40 - 8 nos. (10053 mm˛)
Provide ties #10 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG61 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 688.25 kN
MomentX,(Mx) = 52.77 kN-m
MomentY,(My) = 18.33 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C690.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 14.247 kN.m
My_MinEccen = 13.765 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 52.768 kN.m
My = max(My,My_MinEccen) + MuaddY = 18.333 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 52.77/688.25 = 77 mm
Actual eccenY = My / P = 18.33/688.25 = 27 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(77,21) = 77 mm
eccenY = max(Actual eccenY,eccenYMin) = max(27,20) = 27 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(77 mm) > 0.05 x 300(15 mm)
and eccenY(27 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00292 353.23 8.92 344.31 77.88 179 13.94
2 226 0.00180 317.51 8.83 308.67 69.82 90 6.25
3 226 0.00068 136.48 5.05 131.43 29.73 0 0.00
4 226 -0.00044 -87.68 0.00 -87.68 -19.83 -90 1.78
5 226 -0.00156 -299.99 0.00 -299.99 -67.86 -179 12.15
Total89.74 34.11
xux = 279 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 279/450 = 0.224
Puc = 0.224 x 20.00 x 300 x 450
= 603.70 kN
Pux1 = Puc + Pus(Total)
= 603.70 + (89.74)
= 693.45 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 279/450
= 0.258
Muc = 603.70 x (0.5 x 450 - 0.258 x 450) = 65.64 kN-m
Mux1 = Muc + Mus(Total)
= 65.64 + (34.11)
= 99.75 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00267 349.42 8.92 340.50 192.55 104 20.02
2 565 -0.00110 -219.53 0.00 -219.53 -124.14 -104 12.91
Total68.40 32.94
xuy = 193 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 193/300 = 0.232
Puc = 0.232 x 20.00 x 450 x 300
= 626.48 kN
Puy1 = Puc + Pus(Total)
= 626.48 + (68.40)
= 694.89 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 193/300
= 0.268
Muc = 626.48 x (0.5 x 300 - 0.268 x 300) = 43.58 kN-m
Muy1 = Muc + Mus(Total)
= 43.58 + (32.94)
= 76.52 kN-m
Pu/Puz = 0.442
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.442, an = 1.403
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((52.77/99.75)1.403) + ((18.33/76.52)1.403)
= 0.544 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG61 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1355.84 kN
MomentX,(Mx) = 41.42 kN-m
MomentY,(My) = 18.18 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C690.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 28.066 kN.m
My_MinEccen = 27.117 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 41.418 kN.m
My = max(My,My_MinEccen) + MuaddY = 27.117 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 41.42/1355.84 = 31 mm
Actual eccenY = My / P = 18.18/1355.84 = 13 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(31,21) = 31 mm
eccenY = max(Actual eccenY,eccenYMin) = max(13,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(31 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 14 nos. (1583 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00314 355.09 8.92 346.17 78.30 179 14.02
2 226 0.00267 349.36 8.92 340.44 77.01 119 9.19
3 226 0.00219 334.85 8.92 325.93 73.72 60 4.40
4 226 0.00172 312.46 8.75 303.71 68.70 -0 -0.00
5 226 0.00125 250.16 7.67 242.49 54.85 -60 -3.27
6 226 0.00078 155.87 5.60 150.27 33.99 -119 -4.06
7 226 0.00031 61.58 2.54 59.05 13.36 -179 -2.39
Total399.93 17.88
xux = 443 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 443/450 = 0.354
Puc = 0.354 x 20.00 x 300 x 450
= 956.81 kN
Pux1 = Puc + Pus(Total)
= 956.81 + (399.93)
= 1356.74 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 443/450
= 0.410
Muc = 956.81 x (0.5 x 450 - 0.410 x 450) = 38.97 kN-m
Mux1 = Muc + Mus(Total)
= 38.97 + (17.88)
= 56.85 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 792 0.00291 353.14 8.92 344.22 272.51 104 28.34
2 792 0.00061 122.51 4.63 117.88 93.32 -104 -9.71
Total365.84 18.64
xuy = 309 mm               Puc = C1.fck.B.D
ku = 1.031
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.027
C1 = 0.446 x (1 - C3/6) = 0.370
Puc = 0.370 x 20.00 x 450 x 300
= 998.00 kN
Puy1 = Puc + Pus(Total)
= 998.00 + (365.84)
= 1363.84 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.426
Muc = 998.00 x (0.5 x 300 - 0.426 x 300) = 22.09 kN-m
Muy1 = Muc + Mus(Total)
= 22.09 + (18.64)
= 40.73 kN-m
Pu/Puz = 0.801
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.801, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((41.42/56.85)2.000) + ((27.12/40.73)2.000)
= 0.974 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG61 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 14 nos. (1583 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 2069.10 kN
MomentX,(Mx) = 37.55 kN-m
MomentY,(My) = 17.46 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C690.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 42.830 kN.m
My_MinEccen = 41.382 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 42.830 kN.m
My = max(My,My_MinEccen) + MuaddY = 41.382 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 37.55/2069.10 = 18 mm
Actual eccenY = My / P = 17.46/2069.10 = 8 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(18,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(8,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 14 nos. (4398 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00310 354.82 8.92 345.90 217.33 175 38.03
2 628 0.00264 348.82 8.92 339.90 213.56 117 24.92
3 628 0.00218 334.47 8.92 325.55 204.55 58 11.93
4 628 0.00172 312.46 8.75 303.71 190.83 0 0.00
5 628 0.00126 252.26 7.70 244.56 153.66 -58 -8.96
6 628 0.00080 160.08 5.71 154.37 96.99 -117 -11.32
7 628 0.00034 67.90 2.77 65.13 40.92 -175 -7.16
Total1117.85 47.44
xux = 443 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 443/450 = 0.354
Puc = 0.354 x 20.00 x 300 x 450
= 956.81 kN
Pux1 = Puc + Pus(Total)
= 956.81 + (1117.85)
= 2074.66 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 443/450
= 0.410
Muc = 956.81 x (0.5 x 450 - 0.410 x 450) = 38.97 kN-m
Mux1 = Muc + Mus(Total)
= 38.97 + (47.44)
= 86.41 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 2199 0.00283 352.38 8.92 343.46 755.31 100 75.53
2 2199 0.00072 144.60 5.28 139.32 306.38 -100 -30.64
Total1061.68 44.89
xuy = 319 mm               Puc = C1.fck.B.D
ku = 1.063
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.929
C1 = 0.446 x (1 - C3/6) = 0.377
Puc = 0.377 x 20.00 x 450 x 300
= 1017.83 kN
Puy1 = Puc + Pus(Total)
= 1017.83 + (1061.68)
= 2079.51 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.435
Muc = 1017.83 x (0.5 x 300 - 0.435 x 300) = 19.97 kN-m
Muy1 = Muc + Mus(Total)
= 19.97 + (44.89)
= 64.86 kN-m
Pu/Puz = 0.813
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.813, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((42.83/86.41)2.000) + ((41.38/64.86)2.000)
= 0.653 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 20 = 320 mm
               iii) 300 mm
               Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG61 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 14 nos. (4398 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 2853.63 kN
MomentX,(Mx) = 31.64 kN-m
MomentY,(My) = 16.69 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C690.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 59.070 kN.m
My_MinEccen = 57.073 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 59.070 kN.m
My = max(My,My_MinEccen) + MuaddY = 57.073 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 31.64/2853.63 = 11 mm
Actual eccenY = My / P = 16.69/2853.63 = 6 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(11,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(6,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #32 - 10 nos. (8042 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1608 0.00304 354.24 8.92 345.32 555.45 169 93.87
2 1608 0.00234 340.38 8.92 331.46 533.14 85 45.05
3 1608 0.00165 307.87 8.64 299.23 481.30 0 0.00
4 1608 0.00095 190.70 6.48 184.23 296.33 -85 -25.04
5 1608 0.00026 51.65 2.16 49.50 79.62 -169 -13.46
Total1945.85 100.43
xux = 425 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 425/450 = 0.340
Puc = 0.340 x 20.00 x 300 x 450
= 918.84 kN
Pux1 = Puc + Pus(Total)
= 918.84 + (1945.85)
= 2864.69 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 425/450
= 0.393
Muc = 918.84 x (0.5 x 450 - 0.393 x 450) = 44.14 kN-m
Mux1 = Muc + Mus(Total)
= 44.14 + (100.43)
= 144.57 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 4021 0.00285 352.56 8.92 343.64 1381.85 94 129.89
2 4021 0.00065 130.67 4.88 125.79 505.83 -94 -47.55
Total1887.69 82.35
xuy = 300 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 300/300 = 0.360
Puc = 0.360 x 20.00 x 450 x 300
= 972.00 kN
Puy1 = Puc + Pus(Total)
= 972.00 + (1887.69)
= 2859.69 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 300/300
= 0.416
Muc = 972.00 x (0.5 x 300 - 0.416 x 300) = 24.49 kN-m
Muy1 = Muc + Mus(Total)
= 24.49 + (82.35)
= 106.84 kN-m
Pu/Puz = 0.783
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.783, an = 1.971
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((59.07/144.57)1.971) + ((57.07/106.84)1.971)
= 0.462 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 32/4 = 8 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (8, 6) = 8 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 32 = 512 mm
               iii) 300 mm
               Required spacing = minimum of (300, 512, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG61 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #32 - 10 nos. (8042 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 600 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 3754.29 kN
MomentX,(Mx) = 12.95 kN-m
MomentY,(My) = 10.35 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C690.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 600 = 5.63 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 100.427 kN.m
My_MinEccen = 75.086 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 100.427 kN.m
My = max(My,My_MinEccen) + MuaddY = 75.086 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 12.95/3754.29 = 3 mm
Actual eccenY = My / P = 10.35/3754.29 = 3 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (600/30) = 27 mm
eccenXMin = max(27,20)= 27 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 27 mm
eccenX = max(Actual eccenX,eccenXMin) = max(3,27) = 27 mm
eccenY = max(Actual eccenY,eccenYMin) = max(3,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(27 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 600(30 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #25 - 20 nos. (9817 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 982 0.00319 355.55 8.92 346.63 340.31 248 84.23
2 982 0.00286 352.70 8.92 343.78 337.51 193 64.97
3 982 0.00254 346.07 8.92 337.15 330.99 138 45.51
4 982 0.00221 335.49 8.92 326.57 320.61 83 26.45
5 982 0.00189 322.63 8.89 313.74 308.01 28 8.47
6 982 0.00156 300.00 8.49 291.51 286.19 -28 -7.87
7 982 0.00123 246.67 7.61 239.06 234.69 -83 -19.36
8 982 0.00091 181.48 6.26 175.22 172.03 -138 -23.65
9 982 0.00058 116.30 4.43 111.86 109.82 -193 -21.14
10 982 0.00026 51.11 2.13 48.98 48.08 -248 -11.90
Total2488.24 145.70
xux = 591 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 591/600 = 0.354
Puc = 0.354 x 20.00 x 300 x 600
= 1275.75 kN
Pux1 = Puc + Pus(Total)
= 1275.75 + (2488.24)
= 3763.99 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 591/600
= 0.410
Muc = 1275.75 x (0.5 x 600 - 0.410 x 600) = 69.27 kN-m
Mux1 = Muc + Mus(Total)
= 69.27 + (145.70)
= 214.97 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 4909 0.00278 351.98 8.92 343.06 1684.00 98 164.19
2 4909 0.00078 155.88 5.60 150.28 737.68 -98 -71.92
Total2421.68 92.27
xuy = 323 mm               Puc = C1.fck.B.D
ku = 1.078
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.884
C1 = 0.446 x (1 - C3/6) = 0.380
Puc = 0.380 x 20.00 x 600 x 300
= 1368.91 kN
Puy1 = Puc + Pus(Total)
= 1368.91 + (2421.68)
= 3790.59 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.438
Muc = 1368.91 x (0.5 x 300 - 0.438 x 300) = 25.36 kN-m
Muy1 = Muc + Mus(Total)
= 25.36 + (92.27)
= 117.63 kN-m
Pu/Puz = 0.818
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.818, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((100.43/214.97)2.000) + ((75.09/117.63)2.000)
= 0.626 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 25/4 = 6 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (6, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 25 = 400 mm
               iii) 300 mm
               Required spacing = minimum of (300, 400, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG61 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 600 mm
Provide #25 - 20 nos. (9817 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 6 legged ties along width and
2 legged ties along depth.




Design of CG62 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 365.18 kN
MomentX,(Mx) = 31.55 kN-m
MomentY,(My) = 54.97 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C700.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 7.559 kN.m
My_MinEccen = 7.304 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 31.553 kN.m
My = max(My,My_MinEccen) + MuaddY = 54.965 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 31.55/365.18 = 86 mm
Actual eccenY = My / P = 54.97/365.18 = 151 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(86,21) = 86 mm
eccenY = max(Actual eccenY,eccenYMin) = max(151,20) = 151 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(86 mm) > 0.05 x 300(15 mm)
and eccenY(151 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00268 349.71 8.92 340.79 77.08 179 13.80
2 226 0.00108 216.06 7.03 209.03 47.28 90 4.23
3 226 -0.00052 -103.59 0.00 -103.59 -23.43 0 0.00
4 226 -0.00212 -332.01 0.00 -332.01 -75.10 -90 6.72
5 226 -0.00371 -360.15 0.00 -360.15 -81.46 -179 14.58
Total-55.63 39.33
xux = 196 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 196/450 = 0.157
Puc = 0.157 x 20.00 x 300 x 450
= 423.35 kN
Pux1 = Puc + Pus(Total)
= 423.35 + (-55.63)
= 367.72 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 196/450
= 0.181
Muc = 423.35 x (0.5 x 450 - 0.181 x 450) = 60.74 kN-m
Mux1 = Muc + Mus(Total)
= 60.74 + (39.33)
= 100.07 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00215 333.36 8.92 324.44 183.47 104 19.08
2 565 -0.00394 -360.90 0.00 -360.90 -204.08 -104 21.22
Total-20.62 40.31
xuy = 120 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 120/300 = 0.143
Puc = 0.143 x 20.00 x 450 x 300
= 387.28 kN
Puy1 = Puc + Pus(Total)
= 387.28 + (-20.62)
= 366.66 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 120/300
= 0.166
Muc = 387.28 x (0.5 x 300 - 0.166 x 300) = 38.83 kN-m
Muy1 = Muc + Mus(Total)
= 38.83 + (40.31)
= 79.14 kN-m
Pu/Puz = 0.235
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.235, an = 1.058
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((31.55/100.07)1.058) + ((54.97/79.14)1.058)
= 0.975 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG62 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 728.20 kN
MomentX,(Mx) = 26.72 kN-m
MomentY,(My) = 49.53 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C700.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 15.074 kN.m
My_MinEccen = 14.564 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 26.717 kN.m
My = max(My,My_MinEccen) + MuaddY = 49.531 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 26.72/728.20 = 37 mm
Actual eccenY = My / P = 49.53/728.20 = 68 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(37,21) = 37 mm
eccenY = max(Actual eccenY,eccenYMin) = max(68,20) = 68 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(37 mm) > 0.05 x 300(15 mm)
and eccenY(68 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00294 353.42 8.92 344.50 77.92 179 13.95
2 226 0.00186 321.36 8.88 312.48 70.68 90 6.33
3 226 0.00078 156.97 5.63 151.34 34.23 0 0.00
4 226 -0.00030 -59.04 0.00 -59.04 -13.35 -90 1.20
5 226 -0.00138 -275.04 0.00 -275.04 -62.21 -179 11.14
Total107.27 32.61
xux = 290 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 290/450 = 0.232
Puc = 0.232 x 20.00 x 300 x 450
= 626.48 kN
Pux1 = Puc + Pus(Total)
= 626.48 + (107.27)
= 733.76 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 290/450
= 0.268
Muc = 626.48 x (0.5 x 450 - 0.268 x 450) = 65.37 kN-m
Mux1 = Muc + Mus(Total)
= 65.37 + (32.61)
= 97.98 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00269 350.05 8.92 341.13 192.90 104 20.06
2 565 -0.00096 -192.49 0.00 -192.49 -108.85 -104 11.32
Total84.05 31.38
xuy = 199 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 199/300 = 0.239
Puc = 0.239 x 20.00 x 450 x 300
= 645.47 kN
Puy1 = Puc + Pus(Total)
= 645.47 + (84.05)
= 729.52 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 199/300
= 0.276
Muc = 645.47 x (0.5 x 300 - 0.276 x 300) = 43.33 kN-m
Muy1 = Muc + Mus(Total)
= 43.33 + (31.38)
= 74.71 kN-m
Pu/Puz = 0.468
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.468, an = 1.446
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((26.72/97.98)1.446) + ((49.53/74.71)1.446)
= 0.705 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG62 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1095.07 kN
MomentX,(Mx) = 27.05 kN-m
MomentY,(My) = 44.06 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C700.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 22.668 kN.m
My_MinEccen = 21.901 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 27.049 kN.m
My = max(My,My_MinEccen) + MuaddY = 44.064 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 27.05/1095.07 = 25 mm
Actual eccenY = My / P = 44.06/1095.07 = 40 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(25,21) = 25 mm
eccenY = max(Actual eccenY,eccenYMin) = max(40,20) = 40 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(25 mm) > 0.05 x 300(15 mm)
and eccenY(40 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00309 354.73 8.92 345.81 78.22 179 14.00
2 226 0.00231 338.99 8.92 330.07 74.66 90 6.68
3 226 0.00152 296.06 8.40 287.66 65.07 0 0.00
4 226 0.00073 145.84 5.32 140.52 31.78 -90 -2.84
5 226 -0.00006 -11.87 0.00 -11.87 -2.68 -179 0.48
Total247.05 18.32
xux = 397 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 397/450 = 0.318
Puc = 0.318 x 20.00 x 300 x 450
= 858.09 kN
Pux1 = Puc + Pus(Total)
= 858.09 + (247.05)
= 1105.14 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 397/450
= 0.367
Muc = 858.09 x (0.5 x 450 - 0.367 x 450) = 51.26 kN-m
Mux1 = Muc + Mus(Total)
= 51.26 + (18.32)
= 69.58 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00291 353.09 8.92 344.17 194.63 104 20.24
2 565 0.00023 46.02 1.93 44.09 24.93 -104 -2.59
Total219.56 17.65
xuy = 272 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 272/300 = 0.326
Puc = 0.326 x 20.00 x 450 x 300
= 880.88 kN
Puy1 = Puc + Pus(Total)
= 880.88 + (219.56)
= 1100.43 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 272/300
= 0.377
Muc = 880.88 x (0.5 x 300 - 0.377 x 300) = 32.50 kN-m
Muy1 = Muc + Mus(Total)
= 32.50 + (17.65)
= 50.15 kN-m
Pu/Puz = 0.703
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.703, an = 1.839
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((27.05/69.58)1.839) + ((44.06/50.15)1.839)
= 0.964 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG62 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1465.52 kN
MomentX,(Mx) = 27.79 kN-m
MomentY,(My) = 35.88 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C700.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 30.336 kN.m
My_MinEccen = 29.310 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 30.336 kN.m
My = max(My,My_MinEccen) + MuaddY = 35.877 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 27.79/1465.52 = 19 mm
Actual eccenY = My / P = 35.88/1465.52 = 24 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(19,21) = 21 mm
eccenY = max(Actual eccenY,eccenYMin) = max(24,20) = 24 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(21 mm) > 0.05 x 300(15 mm)
and eccenY(24 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 10 nos. (2011 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00313 355.01 8.92 346.09 139.17 177 24.63
2 402 0.00244 343.53 8.92 334.61 134.55 89 11.91
3 402 0.00175 314.19 8.78 305.41 122.81 0 0.00
4 402 0.00106 212.33 6.96 205.38 82.59 -89 -7.31
5 402 0.00037 74.67 3.02 71.65 28.81 -177 -5.10
Total507.93 24.13
xux = 450 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 450/450 = 0.360
Puc = 0.360 x 20.00 x 300 x 450
= 972.00 kN
Pux1 = Puc + Pus(Total)
= 972.00 + (507.93)
= 1479.93 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 450/450
= 0.416
Muc = 972.00 x (0.5 x 450 - 0.416 x 450) = 36.74 kN-m
Mux1 = Muc + Mus(Total)
= 36.74 + (24.13)
= 60.87 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1005 0.00289 352.95 8.92 344.03 345.86 102 35.28
2 1005 0.00063 126.93 4.76 122.17 122.82 -102 -12.53
Total468.67 22.75
xuy = 309 mm               Puc = C1.fck.B.D
ku = 1.031
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.027
C1 = 0.446 x (1 - C3/6) = 0.370
Puc = 0.370 x 20.00 x 450 x 300
= 998.00 kN
Puy1 = Puc + Pus(Total)
= 998.00 + (468.67)
= 1466.67 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.426
Muc = 998.00 x (0.5 x 300 - 0.426 x 300) = 22.09 kN-m
Muy1 = Muc + Mus(Total)
= 22.09 + (22.75)
= 44.84 kN-m
Pu/Puz = 0.804
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.804, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((30.34/60.87)2.000) + ((35.88/44.84)2.000)
= 0.888 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG62 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 10 nos. (2011 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 1850.52 kN
MomentX,(Mx) = 17.36 kN-m
MomentY,(My) = 19.09 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C700.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 40.249 kN.m
My_MinEccen = 37.010 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 40.249 kN.m
My = max(My,My_MinEccen) + MuaddY = 37.010 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 17.36/1850.52 = 9 mm
Actual eccenY = My / P = 19.09/1850.52 = 10 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(9,22) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(10,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 18 nos. (3619 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00311 354.90 8.92 345.98 139.13 177 24.63
2 402 0.00276 351.78 8.92 342.86 137.87 133 18.30
3 402 0.00240 342.58 8.92 333.66 134.17 89 11.87
4 402 0.00205 329.53 8.92 320.61 128.93 44 5.70
5 402 0.00169 310.67 8.71 301.96 121.42 0 0.00
6 402 0.00134 267.66 7.94 259.71 104.44 -44 -4.62
7 402 0.00098 196.60 6.61 189.99 76.40 -89 -6.76
8 402 0.00063 125.55 4.72 120.83 48.59 -133 -6.45
9 402 0.00027 54.49 2.26 52.23 21.00 -177 -3.72
Total911.95 38.96
xux = 436 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 436/450 = 0.349
Puc = 0.349 x 20.00 x 300 x 450
= 941.63 kN
Pux1 = Puc + Pus(Total)
= 941.63 + (911.95)
= 1853.58 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 436/450
= 0.403
Muc = 941.63 x (0.5 x 450 - 0.403 x 450) = 41.10 kN-m
Mux1 = Muc + Mus(Total)
= 41.10 + (38.96)
= 80.06 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1810 0.00287 352.75 8.92 343.83 622.18 102 63.46
2 1810 0.00067 133.83 4.97 128.86 233.19 -102 -23.78
Total855.37 39.68
xuy = 314 mm               Puc = C1.fck.B.D
ku = 1.047
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.976
C1 = 0.446 x (1 - C3/6) = 0.373
Puc = 0.373 x 20.00 x 450 x 300
= 1008.29 kN
Puy1 = Puc + Pus(Total)
= 1008.29 + (855.37)
= 1863.66 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.431
Muc = 1008.29 x (0.5 x 300 - 0.431 x 300) = 20.99 kN-m
Muy1 = Muc + Mus(Total)
= 20.99 + (39.68)
= 60.67 kN-m
Pu/Puz = 0.801
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.801, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((40.25/80.06)2.000) + ((37.01/60.67)2.000)
= 0.625 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG62 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 18 nos. (3619 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG63 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 370.59 kN
MomentX,(Mx) = 61.51 kN-m
MomentY,(My) = 100.07 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C710.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 7.671 kN.m
My_MinEccen = 7.412 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 61.508 kN.m
My = max(My,My_MinEccen) + MuaddY = 100.068 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 61.51/370.59 = 166 mm
Actual eccenY = My / P = 100.07/370.59 = 270 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(166,21) = 166 mm
eccenY = max(Actual eccenY,eccenYMin) = max(270,20) = 270 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(166 mm) > 0.05 x 300(15 mm)
and eccenY(270 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 18 nos. (3619 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00271 350.56 8.92 341.64 137.38 177 24.32
2 402 0.00199 327.19 8.92 318.27 127.98 133 16.99
3 402 0.00126 251.69 7.69 244.00 98.12 89 8.68
4 402 0.00053 106.36 4.11 102.25 41.12 44 1.82
5 402 -0.00019 -38.97 0.00 -38.97 -15.67 0 0.00
6 402 -0.00092 -184.30 0.00 -184.30 -74.11 -44 3.28
7 402 -0.00165 -307.83 0.00 -307.83 -123.79 -89 10.96
8 402 -0.00237 -341.51 0.00 -341.51 -137.33 -133 18.23
9 402 -0.00310 -354.79 0.00 -354.79 -142.67 -177 25.25
Total-88.96 109.53
xux = 213 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 213/450 = 0.171
Puc = 0.171 x 20.00 x 300 x 450
= 460.37 kN
Pux1 = Puc + Pus(Total)
= 460.37 + (-88.96)
= 371.41 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 213/450
= 0.197
Muc = 460.37 x (0.5 x 450 - 0.197 x 450) = 62.77 kN-m
Mux1 = Muc + Mus(Total)
= 62.77 + (109.53)
= 172.29 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1810 0.00222 335.82 8.92 326.90 591.55 102 60.34
2 1810 -0.00322 -355.83 0.00 -355.83 -643.89 -102 65.68
Total-52.34 126.01
xuy = 131 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 131/300 = 0.157
Puc = 0.157 x 20.00 x 450 x 300
= 425.25 kN
Puy1 = Puc + Pus(Total)
= 425.25 + (-52.34)
= 372.91 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 131/300
= 0.182
Muc = 425.25 x (0.5 x 300 - 0.182 x 300) = 40.57 kN-m
Muy1 = Muc + Mus(Total)
= 40.57 + (126.01)
= 166.58 kN-m
Pu/Puz = 0.161
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.161, an = 1.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((61.51/172.29)1.000) + ((100.07/166.58)1.000)
= 0.958 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG63 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 18 nos. (3619 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 766.70 kN
MomentX,(Mx) = 54.08 kN-m
MomentY,(My) = 90.09 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C710.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 15.871 kN.m
My_MinEccen = 15.334 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 54.075 kN.m
My = max(My,My_MinEccen) + MuaddY = 90.091 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 54.08/766.70 = 71 mm
Actual eccenY = My / P = 90.09/766.70 = 118 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(71,21) = 71 mm
eccenY = max(Actual eccenY,eccenYMin) = max(118,20) = 118 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(71 mm) > 0.05 x 300(15 mm)
and eccenY(118 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 14 nos. (2815 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00287 352.77 8.92 343.85 138.27 177 24.47
2 402 0.00210 331.35 8.92 322.43 129.66 118 15.30
3 402 0.00133 265.10 7.91 257.19 103.42 59 6.10
4 402 0.00055 110.53 4.25 106.28 42.74 0 0.00
5 402 -0.00022 -44.05 0.00 -44.05 -17.71 -59 1.05
6 402 -0.00099 -198.62 0.00 -198.62 -79.87 -118 9.42
7 402 -0.00177 -315.19 0.00 -315.19 -126.74 -177 22.43
Total189.76 78.78
xux = 267 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 267/450 = 0.214
Puc = 0.214 x 20.00 x 300 x 450
= 577.13 kN
Pux1 = Puc + Pus(Total)
= 577.13 + (189.76)
= 766.89 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 267/450
= 0.247
Muc = 577.13 x (0.5 x 450 - 0.247 x 450) = 65.71 kN-m
Mux1 = Muc + Mus(Total)
= 65.71 + (78.78)
= 144.48 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1407 0.00262 348.07 8.92 339.15 477.34 102 48.69
2 1407 -0.00115 -229.19 0.00 -229.19 -322.56 -102 32.90
Total154.77 81.59
xuy = 190 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 190/300 = 0.228
Puc = 0.228 x 20.00 x 450 x 300
= 615.09 kN
Puy1 = Puc + Pus(Total)
= 615.09 + (154.77)
= 769.87 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 190/300
= 0.263
Muc = 615.09 x (0.5 x 300 - 0.263 x 300) = 43.69 kN-m
Muy1 = Muc + Mus(Total)
= 43.69 + (81.59)
= 125.28 kN-m
Pu/Puz = 0.371
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.371, an = 1.285
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((54.08/144.48)1.285) + ((90.09/125.28)1.285)
= 0.937 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG63 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 14 nos. (2815 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1156.85 kN
MomentX,(Mx) = 50.80 kN-m
MomentY,(My) = 85.25 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C710.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 23.947 kN.m
My_MinEccen = 23.137 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 50.805 kN.m
My = max(My,My_MinEccen) + MuaddY = 85.249 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 50.80/1156.85 = 44 mm
Actual eccenY = My / P = 85.25/1156.85 = 74 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(44,21) = 44 mm
eccenY = max(Actual eccenY,eccenYMin) = max(74,20) = 74 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(44 mm) > 0.05 x 300(15 mm)
and eccenY(74 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 16 nos. (3217 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00297 353.68 8.92 344.76 138.64 177 24.54
2 402 0.00242 343.10 8.92 334.18 134.38 126 16.99
3 402 0.00187 321.58 8.88 312.70 125.74 76 9.54
4 402 0.00132 263.02 7.87 255.14 102.60 25 2.59
5 402 0.00076 152.37 5.50 146.87 59.06 -25 -1.49
6 402 0.00021 41.71 1.76 39.95 16.07 -76 -1.22
7 402 -0.00034 -68.94 0.00 -68.94 -27.72 -126 3.50
8 402 -0.00090 -179.59 0.00 -179.59 -72.22 -177 12.78
Total476.54 67.24
xux = 320 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 320/450 = 0.256
Puc = 0.256 x 20.00 x 300 x 450
= 691.03 kN
Pux1 = Puc + Pus(Total)
= 691.03 + (476.54)
= 1167.57 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 320/450
= 0.296
Muc = 691.03 x (0.5 x 450 - 0.296 x 450) = 63.51 kN-m
Mux1 = Muc + Mus(Total)
= 63.51 + (67.24)
= 130.75 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1608 0.00276 351.81 8.92 342.89 551.54 102 56.26
2 1608 -0.00038 -75.92 0.00 -75.92 -122.11 -102 12.46
Total429.42 68.71
xuy = 227 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 227/300 = 0.273
Puc = 0.273 x 20.00 x 450 x 300
= 736.59 kN
Puy1 = Puc + Pus(Total)
= 736.59 + (429.42)
= 1166.02 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 227/300
= 0.315
Muc = 736.59 x (0.5 x 300 - 0.315 x 300) = 40.83 kN-m
Muy1 = Muc + Mus(Total)
= 40.83 + (68.71)
= 109.54 kN-m
Pu/Puz = 0.529
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.529, an = 1.548
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((50.80/130.75)1.548) + ((85.25/109.54)1.548)
= 0.910 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG63 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 16 nos. (3217 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1538.17 kN
MomentX,(Mx) = 45.64 kN-m
MomentY,(My) = 78.42 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C710.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 31.840 kN.m
My_MinEccen = 30.763 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 45.635 kN.m
My = max(My,My_MinEccen) + MuaddY = 78.419 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 45.64/1538.17 = 30 mm
Actual eccenY = My / P = 78.42/1538.17 = 51 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(30,21) = 30 mm
eccenY = max(Actual eccenY,eccenYMin) = max(51,20) = 51 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(30 mm) > 0.05 x 300(15 mm)
and eccenY(51 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 18 nos. (3619 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00305 354.33 8.92 345.41 138.90 177 24.58
2 402 0.00263 348.55 8.92 339.63 136.57 133 18.13
3 402 0.00222 335.75 8.92 326.83 131.42 89 11.63
4 402 0.00180 317.46 8.83 308.63 124.11 44 5.49
5 402 0.00139 277.36 8.08 269.28 108.28 0 0.00
6 402 0.00097 194.24 6.56 187.68 75.47 -44 -3.34
7 402 0.00056 111.12 4.27 106.85 42.97 -89 -3.80
8 402 0.00014 28.00 1.21 26.79 10.77 -133 -1.43
9 402 -0.00028 -55.12 0.00 -55.12 -22.16 -177 3.92
Total746.33 55.19
xux = 373 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 373/450 = 0.298
Puc = 0.298 x 20.00 x 300 x 450
= 804.94 kN
Pux1 = Puc + Pus(Total)
= 804.94 + (746.33)
= 1551.27 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 373/450
= 0.345
Muc = 804.94 x (0.5 x 450 - 0.345 x 450) = 56.33 kN-m
Mux1 = Muc + Mus(Total)
= 56.33 + (55.19)
= 111.51 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1810 0.00287 352.72 8.92 343.80 622.13 102 63.46
2 1810 0.00017 33.95 1.45 32.50 58.81 -102 -6.00
Total680.94 57.46
xuy = 265 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 265/300 = 0.318
Puc = 0.318 x 20.00 x 450 x 300
= 858.09 kN
Puy1 = Puc + Pus(Total)
= 858.09 + (680.94)
= 1539.03 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 265/300
= 0.367
Muc = 858.09 x (0.5 x 300 - 0.367 x 300) = 34.17 kN-m
Muy1 = Muc + Mus(Total)
= 34.17 + (57.46)
= 91.63 kN-m
Pu/Puz = 0.666
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.666, an = 1.777
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((45.64/111.51)1.777) + ((78.42/91.63)1.777)
= 0.963 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG63 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 18 nos. (3619 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 1903.55 kN
MomentX,(Mx) = 28.91 kN-m
MomentY,(My) = 51.44 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C710.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 41.402 kN.m
My_MinEccen = 38.071 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 41.402 kN.m
My = max(My,My_MinEccen) + MuaddY = 51.445 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 28.91/1903.55 = 15 mm
Actual eccenY = My / P = 51.44/1903.55 = 27 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(15,22) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(27,20) = 27 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(27 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 14 nos. (4398 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00307 354.55 8.92 345.63 217.17 175 38.00
2 628 0.00258 347.12 8.92 338.20 212.50 117 24.79
3 628 0.00208 330.74 8.92 321.82 202.21 58 11.80
4 628 0.00159 302.48 8.54 293.94 184.69 0 0.00
5 628 0.00109 217.82 7.07 210.75 132.42 -58 -7.72
6 628 0.00059 118.55 4.50 114.05 71.66 -117 -8.36
7 628 0.00010 19.28 0.84 18.44 11.59 -175 -2.03
Total1032.23 56.48
xux = 411 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 411/450 = 0.329
Puc = 0.329 x 20.00 x 300 x 450
= 888.47 kN
Pux1 = Puc + Pus(Total)
= 888.47 + (1032.23)
= 1920.70 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 411/450
= 0.380
Muc = 888.47 x (0.5 x 450 - 0.380 x 450) = 47.88 kN-m
Mux1 = Muc + Mus(Total)
= 47.88 + (56.48)
= 104.36 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 2199 0.00290 353.01 8.92 344.09 756.68 100 75.67
2 2199 0.00049 97.85 3.83 94.02 206.76 -100 -20.68
Total963.44 54.99
xuy = 291 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 291/300 = 0.349
Puc = 0.349 x 20.00 x 450 x 300
= 941.63 kN
Puy1 = Puc + Pus(Total)
= 941.63 + (963.44)
= 1905.07 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 291/300
= 0.403
Muc = 941.63 x (0.5 x 300 - 0.403 x 300) = 27.40 kN-m
Muy1 = Muc + Mus(Total)
= 27.40 + (54.99)
= 82.39 kN-m
Pu/Puz = 0.748
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.748, an = 1.914
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((41.40/104.36)1.914) + ((51.44/82.39)1.914)
= 0.577 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 20 = 320 mm
               iii) 300 mm
               Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG63 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 14 nos. (4398 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG64 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 779.20 kN
MomentX,(Mx) = 124.49 kN-m
MomentY,(My) = 19.60 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C720.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 16.129 kN.m
My_MinEccen = 15.584 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 124.491 kN.m
My = max(My,My_MinEccen) + MuaddY = 19.596 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 124.49/779.20 = 160 mm
Actual eccenY = My / P = 19.60/779.20 = 25 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(160,21) = 160 mm
eccenY = max(Actual eccenY,eccenYMin) = max(25,20) = 25 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(160 mm) > 0.05 x 300(15 mm)
and eccenY(25 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 16 nos. + #16 - 4 nos. (2614 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00289 352.91 8.92 343.99 138.33 177 24.48
2 226 0.00237 341.16 8.92 332.24 75.15 136 10.23
3 226 0.00187 321.62 8.88 312.74 70.74 97 6.88
4 226 0.00137 274.55 8.04 266.51 60.28 58 3.52
5 226 0.00088 175.28 6.10 169.17 38.27 19 0.74
6 226 0.00038 76.01 3.07 72.94 16.50 -19 -0.32
7 226 -0.00012 -23.27 0.00 -23.27 -5.26 -58 0.31
8 226 -0.00061 -122.54 0.00 -122.54 -27.72 -97 2.69
9 226 -0.00111 -221.81 0.00 -221.81 -50.17 -136 6.83
10 402 -0.00163 -306.76 0.00 -306.76 -123.35 -177 21.83
Total192.76 77.20
xux = 274 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 274/450 = 0.219
Puc = 0.219 x 20.00 x 300 x 450
= 592.31 kN
Pux1 = Puc + Pus(Total)
= 592.31 + (192.76)
= 785.07 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 274/450
= 0.254
Muc = 592.31 x (0.5 x 450 - 0.254 x 450) = 65.70 kN-m
Mux1 = Muc + Mus(Total)
= 65.70 + (77.20)
= 142.90 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1307 0.00263 348.35 8.92 339.43 443.60 102 45.25
2 1307 -0.00109 -217.85 0.00 -217.85 -284.71 -102 29.04
Total158.89 74.29
xuy = 192 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 192/300 = 0.231
Puc = 0.231 x 20.00 x 450 x 300
= 622.69 kN
Puy1 = Puc + Pus(Total)
= 622.69 + (158.89)
= 781.58 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 192/300
= 0.266
Muc = 622.69 x (0.5 x 300 - 0.266 x 300) = 43.62 kN-m
Muy1 = Muc + Mus(Total)
= 43.62 + (74.29)
= 117.91 kN-m
Pu/Puz = 0.389
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.389, an = 1.314
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((124.49/142.90)1.314) + ((19.60/117.91)1.314)
= 0.929 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG64 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 16 nos. + #16 - 4 nos. (2614 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1564.98 kN
MomentX,(Mx) = 107.81 kN-m
MomentY,(My) = 17.73 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C720.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 32.395 kN.m
My_MinEccen = 31.300 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 107.811 kN.m
My = max(My,My_MinEccen) + MuaddY = 31.300 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 107.81/1564.98 = 69 mm
Actual eccenY = My / P = 17.73/1564.98 = 11 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(69,21) = 69 mm
eccenY = max(Actual eccenY,eccenYMin) = max(11,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(69 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 12 nos. (3770 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00303 354.17 8.92 345.25 216.92 175 37.96
2 628 0.00237 341.44 8.92 332.52 208.93 105 21.94
3 628 0.00172 312.04 8.74 303.30 190.57 35 6.67
4 628 0.00106 211.61 6.94 204.67 128.60 -35 -4.50
5 628 0.00040 80.13 3.22 76.91 48.32 -105 -5.07
6 628 -0.00026 -51.36 0.00 -51.36 -32.27 -175 5.65
Total761.07 62.64
xux = 373 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 373/450 = 0.298
Puc = 0.298 x 20.00 x 300 x 450
= 804.94 kN
Pux1 = Puc + Pus(Total)
= 804.94 + (761.07)
= 1566.01 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 373/450
= 0.345
Muc = 804.94 x (0.5 x 450 - 0.345 x 450) = 56.33 kN-m
Mux1 = Muc + Mus(Total)
= 56.33 + (62.64)
= 118.97 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1885 0.00284 352.49 8.92 343.57 647.62 100 64.76
2 1885 0.00020 39.23 1.66 37.57 70.82 -100 -7.08
Total718.44 57.68
xuy = 265 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 265/300 = 0.318
Puc = 0.318 x 20.00 x 450 x 300
= 858.09 kN
Puy1 = Puc + Pus(Total)
= 858.09 + (718.44)
= 1576.53 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 265/300
= 0.367
Muc = 858.09 x (0.5 x 300 - 0.367 x 300) = 34.17 kN-m
Muy1 = Muc + Mus(Total)
= 34.17 + (57.68)
= 91.85 kN-m
Pu/Puz = 0.665
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.665, an = 1.774
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((107.81/118.97)1.774) + ((31.30/91.85)1.774)
= 0.988 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 20 = 320 mm
               iii) 300 mm
               Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG64 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 12 nos. (3770 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 2347.55 kN
MomentX,(Mx) = 103.39 kN-m
MomentY,(My) = 20.28 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C720.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 48.594 kN.m
My_MinEccen = 46.951 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 103.387 kN.m
My = max(My,My_MinEccen) + MuaddY = 46.951 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 103.39/2347.55 = 44 mm
Actual eccenY = My / P = 20.28/2347.55 = 9 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(44,21) = 44 mm
eccenY = max(Actual eccenY,eccenYMin) = max(9,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(44 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #25 - 14 nos. (6872 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 982 0.00303 354.15 8.92 345.23 338.93 173 58.47
2 982 0.00251 345.46 8.92 336.54 330.40 115 38.00
3 982 0.00200 327.65 8.92 318.73 312.92 58 17.99
4 982 0.00148 292.68 8.32 284.36 279.17 0 0.00
5 982 0.00097 193.25 6.54 186.72 183.31 -58 -10.54
6 982 0.00045 90.11 3.57 86.54 84.96 -115 -9.77
7 982 -0.00007 -13.03 0.00 -13.03 -12.80 -173 2.21
Total1516.89 96.35
xux = 390 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 390/450 = 0.312
Puc = 0.312 x 20.00 x 300 x 450
= 842.91 kN
Pux1 = Puc + Pus(Total)
= 842.91 + (1516.89)
= 2359.80 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 390/450
= 0.361
Muc = 842.91 x (0.5 x 450 - 0.361 x 450) = 52.82 kN-m
Mux1 = Muc + Mus(Total)
= 52.82 + (96.35)
= 149.17 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 3436 0.00285 352.56 8.92 343.64 1180.78 98 115.13
2 3436 0.00042 84.00 3.35 80.65 277.11 -98 -27.02
Total1457.89 88.11
xuy = 281 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 281/300 = 0.338
Puc = 0.338 x 20.00 x 450 x 300
= 911.25 kN
Puy1 = Puc + Pus(Total)
= 911.25 + (1457.89)
= 2369.14 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 281/300
= 0.390
Muc = 911.25 x (0.5 x 300 - 0.390 x 300) = 30.07 kN-m
Muy1 = Muc + Mus(Total)
= 30.07 + (88.11)
= 118.18 kN-m
Pu/Puz = 0.713
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.713, an = 1.855
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((103.39/149.17)1.855) + ((46.95/118.18)1.855)
= 0.687 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 25/4 = 6 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (6, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 25 = 400 mm
               iii) 300 mm
               Required spacing = minimum of (300, 400, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG64 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #25 - 14 nos. (6872 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 3124.29 kN
MomentX,(Mx) = 96.50 kN-m
MomentY,(My) = 24.85 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C720.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 64.673 kN.m
My_MinEccen = 62.486 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 96.495 kN.m
My = max(My,My_MinEccen) + MuaddY = 62.486 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 96.50/3124.29 = 31 mm
Actual eccenY = My / P = 24.85/3124.29 = 8 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(31,21) = 31 mm
eccenY = max(Actual eccenY,eccenYMin) = max(8,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(31 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #32 - 12 nos. (9651 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1608 0.00302 354.07 8.92 345.15 555.17 169 93.82
2 1608 0.00244 343.55 8.92 334.63 538.25 101 54.58
3 1608 0.00186 321.00 8.88 312.12 502.04 34 16.97
4 1608 0.00128 255.78 7.76 248.02 398.93 -34 -13.48
5 1608 0.00070 139.74 5.14 134.60 216.50 -101 -21.95
6 1608 0.00012 23.71 1.03 22.68 36.49 -169 -6.17
Total2247.39 123.77
xux = 408 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 408/450 = 0.326
Puc = 0.326 x 20.00 x 300 x 450
= 880.88 kN
Pux1 = Puc + Pus(Total)
= 880.88 + (2247.39)
= 3128.26 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 408/450
= 0.377
Muc = 880.88 x (0.5 x 450 - 0.377 x 450) = 48.76 kN-m
Mux1 = Muc + Mus(Total)
= 48.76 + (123.77)
= 172.53 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 4825 0.00283 352.42 8.92 343.50 1657.56 94 155.81
2 4825 0.00059 117.00 4.46 112.55 543.10 -94 -51.05
Total2200.66 104.76
xuy = 293 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 293/300 = 0.352
Puc = 0.352 x 20.00 x 450 x 300
= 949.22 kN
Puy1 = Puc + Pus(Total)
= 949.22 + (2200.66)
= 3149.88 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 293/300
= 0.406
Muc = 949.22 x (0.5 x 300 - 0.406 x 300) = 26.70 kN-m
Muy1 = Muc + Mus(Total)
= 26.70 + (104.76)
= 131.46 kN-m
Pu/Puz = 0.756
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.756, an = 1.927
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((96.50/172.53)1.927) + ((62.49/131.46)1.927)
= 0.565 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 32/4 = 8 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (8, 6) = 8 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 32 = 512 mm
               iii) 300 mm
               Required spacing = minimum of (300, 512, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG64 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #32 - 12 nos. (9651 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 600 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 3889.92 kN
MomentX,(Mx) = 63.89 kN-m
MomentY,(My) = 16.72 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C720.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 600 = 5.63 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 104.055 kN.m
My_MinEccen = 77.798 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 104.055 kN.m
My = max(My,My_MinEccen) + MuaddY = 77.798 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 63.89/3889.92 = 16 mm
Actual eccenY = My / P = 16.72/3889.92 = 4 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (600/30) = 27 mm
eccenXMin = max(27,20)= 27 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 27 mm
eccenX = max(Actual eccenX,eccenXMin) = max(16,27) = 27 mm
eccenY = max(Actual eccenY,eccenYMin) = max(4,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(27 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 600(30 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #25 - 20 nos. (9817 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 982 0.00313 355.05 8.92 346.13 339.82 248 84.10
2 982 0.00283 352.39 8.92 343.47 337.20 193 64.91
3 982 0.00252 345.72 8.92 336.80 330.65 138 45.46
4 982 0.00222 335.79 8.92 326.87 320.91 83 26.47
5 982 0.00192 324.49 8.90 315.59 309.83 28 8.52
6 982 0.00161 304.89 8.58 296.30 290.90 -28 -8.00
7 982 0.00131 261.33 7.85 253.49 248.86 -83 -20.53
8 982 0.00100 200.49 6.70 193.79 190.26 -138 -26.16
9 982 0.00070 139.65 5.14 134.51 132.06 -193 -25.42
10 982 0.00039 78.81 3.17 75.65 74.27 -248 -18.38
Total2574.74 130.98
xux = 619 mm               Puc = C1.fck.B.D
ku = 1.031
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.027
C1 = 0.446 x (1 - C3/6) = 0.370
Puc = 0.370 x 20.00 x 300 x 600
= 1330.66 kN
Pux1 = Puc + Pus(Total)
= 1330.66 + (2574.74)
= 3905.40 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.426
Muc = 1330.66 x (0.5 x 600 - 0.426 x 600) = 58.91 kN-m
Mux1 = Muc + Mus(Total)
= 58.91 + (130.98)
= 189.90 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 4909 0.00273 350.98 8.92 342.06 1679.10 98 163.71
2 4909 0.00086 172.31 6.03 166.28 816.22 -98 -79.58
Total2495.31 84.13
xuy = 338 mm               Puc = C1.fck.B.D
ku = 1.125
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.769
C1 = 0.446 x (1 - C3/6) = 0.389
Puc = 0.389 x 20.00 x 600 x 300
= 1399.70 kN
Puy1 = Puc + Pus(Total)
= 1399.70 + (2495.31)
= 3895.01 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.447
Muc = 1399.70 x (0.5 x 300 - 0.447 x 300) = 22.06 kN-m
Muy1 = Muc + Mus(Total)
= 22.06 + (84.13)
= 106.19 kN-m
Pu/Puz = 0.848
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.848, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((104.06/189.90)2.000) + ((77.80/106.19)2.000)
= 0.837 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 25/4 = 6 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (6, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 25 = 400 mm
               iii) 300 mm
               Required spacing = minimum of (300, 400, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG64 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 600 mm
Provide #25 - 20 nos. (9817 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 6 legged ties along width and
2 legged ties along depth.




Design of CG65 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 709.65 kN
MomentX,(Mx) = 91.56 kN-m
MomentY,(My) = 0.30 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C730.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 14.690 kN.m
My_MinEccen = 14.193 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 91.562 kN.m
My = max(My,My_MinEccen) + MuaddY = 14.193 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 91.56/709.65 = 129 mm
Actual eccenY = My / P = 0.30/709.65 = 0 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(129,21) = 129 mm
eccenY = max(Actual eccenY,eccenYMin) = max(0,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(129 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00293 353.33 8.92 344.41 77.90 179 13.94
2 226 0.00183 319.47 8.86 310.61 70.26 90 6.29
3 226 0.00073 146.91 5.35 141.56 32.02 0 0.00
4 226 -0.00037 -73.09 0.00 -73.09 -16.53 -90 1.48
5 226 -0.00147 -291.11 0.00 -291.11 -65.85 -179 11.79
Total97.80 33.50
xux = 285 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 285/450 = 0.228
Puc = 0.228 x 20.00 x 300 x 450
= 615.09 kN
Pux1 = Puc + Pus(Total)
= 615.09 + (97.80)
= 712.90 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 285/450
= 0.263
Muc = 615.09 x (0.5 x 450 - 0.263 x 450) = 65.53 kN-m
Mux1 = Muc + Mus(Total)
= 65.53 + (33.50)
= 99.03 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00268 349.80 8.92 340.88 192.76 104 20.05
2 565 -0.00102 -203.11 0.00 -203.11 -114.86 -104 11.95
Total77.91 31.99
xuy = 197 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 197/300 = 0.236
Puc = 0.236 x 20.00 x 450 x 300
= 637.88 kN
Puy1 = Puc + Pus(Total)
= 637.88 + (77.91)
= 715.78 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 197/300
= 0.273
Muc = 637.88 x (0.5 x 300 - 0.273 x 300) = 43.44 kN-m
Muy1 = Muc + Mus(Total)
= 43.44 + (31.99)
= 75.43 kN-m
Pu/Puz = 0.456
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.456, an = 1.426
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((91.56/99.03)1.426) + ((14.19/75.43)1.426)
= 0.986 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG65 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1436.66 kN
MomentX,(Mx) = 79.20 kN-m
MomentY,(My) = 0.44 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C730.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 29.739 kN.m
My_MinEccen = 28.733 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 79.200 kN.m
My = max(My,My_MinEccen) + MuaddY = 28.733 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 79.20/1436.66 = 55 mm
Actual eccenY = My / P = 0.44/1436.66 = 0 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(55,21) = 55 mm
eccenY = max(Actual eccenY,eccenYMin) = max(0,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(55 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 16 nos. + #16 - 4 nos. (2614 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00308 354.61 8.92 345.69 139.01 177 24.60
2 226 0.00272 350.87 8.92 341.95 77.35 136 10.53
3 226 0.00238 341.85 8.92 332.93 75.31 97 7.32
4 226 0.00204 329.37 8.92 320.45 72.48 58 4.23
5 226 0.00170 311.37 8.73 302.65 68.46 19 1.33
6 226 0.00137 273.06 8.02 265.03 59.95 -19 -1.17
7 226 0.00103 205.13 6.80 198.33 44.86 -58 -2.62
8 226 0.00069 137.21 5.07 132.14 29.89 -97 -2.91
9 226 0.00035 69.29 2.82 66.46 15.03 -136 -2.05
10 402 -0.00001 -2.13 0.00 -2.13 -0.86 -177 0.15
Total581.48 39.43
xux = 401 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 401/450 = 0.321
Puc = 0.321 x 20.00 x 300 x 450
= 865.69 kN
Pux1 = Puc + Pus(Total)
= 865.69 + (581.48)
= 1447.17 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 401/450
= 0.370
Muc = 865.69 x (0.5 x 450 - 0.370 x 450) = 50.45 kN-m
Mux1 = Muc + Mus(Total)
= 50.45 + (39.43)
= 89.88 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1307 0.00290 353.00 8.92 344.08 449.68 102 45.87
2 1307 0.00034 67.53 2.76 64.77 84.65 -102 -8.63
Total534.34 37.23
xuy = 279 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 279/300 = 0.335
Puc = 0.335 x 20.00 x 450 x 300
= 903.66 kN
Puy1 = Puc + Pus(Total)
= 903.66 + (534.34)
= 1437.99 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 279/300
= 0.387
Muc = 903.66 x (0.5 x 300 - 0.387 x 300) = 30.70 kN-m
Muy1 = Muc + Mus(Total)
= 30.70 + (37.23)
= 67.94 kN-m
Pu/Puz = 0.717
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.717, an = 1.861
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((79.20/89.88)1.861) + ((28.73/67.94)1.861)
= 0.992 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG65 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 16 nos. + #16 - 4 nos. (2614 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 2149.99 kN
MomentX,(Mx) = 75.89 kN-m
MomentY,(My) = 0.92 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C730.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 44.505 kN.m
My_MinEccen = 43.000 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 75.895 kN.m
My = max(My,My_MinEccen) + MuaddY = 43.000 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 75.89/2149.99 = 35 mm
Actual eccenY = My / P = 0.92/2149.99 = 0 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(35,21) = 35 mm
eccenY = max(Actual eccenY,eccenYMin) = max(0,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(35 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 16 nos. (5027 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00309 354.68 8.92 345.76 217.24 175 38.02
2 628 0.00268 349.67 8.92 340.75 214.10 125 26.76
3 628 0.00227 337.50 8.92 328.58 206.46 75 15.48
4 628 0.00185 320.71 8.87 311.84 195.93 25 4.90
5 628 0.00144 288.99 8.23 280.76 176.41 -25 -4.41
6 628 0.00103 206.34 6.83 199.51 125.35 -75 -9.40
7 628 0.00062 124.06 4.67 119.38 75.01 -125 -9.38
8 628 0.00021 41.78 1.77 40.02 25.14 -175 -4.40
Total1235.65 57.57
xux = 425 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 425/450 = 0.340
Puc = 0.340 x 20.00 x 300 x 450
= 918.84 kN
Pux1 = Puc + Pus(Total)
= 918.84 + (1235.65)
= 2154.49 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 425/450
= 0.393
Muc = 918.84 x (0.5 x 450 - 0.393 x 450) = 44.14 kN-m
Mux1 = Muc + Mus(Total)
= 44.14 + (57.57)
= 101.71 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 2513 0.00289 352.96 8.92 344.04 864.66 100 86.47
2 2513 0.00062 124.21 4.68 119.53 300.41 -100 -30.04
Total1165.07 56.43
xuy = 305 mm               Puc = C1.fck.B.D
ku = 1.016
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.083
C1 = 0.446 x (1 - C3/6) = 0.366
Puc = 0.366 x 20.00 x 450 x 300
= 986.88 kN
Puy1 = Puc + Pus(Total)
= 986.88 + (1165.07)
= 2151.94 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.421
Muc = 986.88 x (0.5 x 300 - 0.421 x 300) = 23.28 kN-m
Muy1 = Muc + Mus(Total)
= 23.28 + (56.43)
= 79.71 kN-m
Pu/Puz = 0.786
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.786, an = 1.977
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((75.89/101.71)1.977) + ((43.00/79.71)1.977)
= 0.856 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 20 = 320 mm
               iii) 300 mm
               Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG65 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 16 nos. (5027 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 2842.53 kN
MomentX,(Mx) = 71.24 kN-m
MomentY,(My) = 1.56 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C730.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 58.840 kN.m
My_MinEccen = 56.851 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 71.242 kN.m
My = max(My,My_MinEccen) + MuaddY = 56.851 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 71.24/2842.53 = 25 mm
Actual eccenY = My / P = 1.56/2842.53 = 1 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(25,21) = 25 mm
eccenY = max(Actual eccenY,eccenYMin) = max(1,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(25 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #32 - 10 nos. (8042 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1608 0.00304 354.24 8.92 345.32 555.45 169 93.87
2 1608 0.00234 340.38 8.92 331.46 533.14 85 45.05
3 1608 0.00165 307.87 8.64 299.23 481.30 0 0.00
4 1608 0.00095 190.70 6.48 184.23 296.33 -85 -25.04
5 1608 0.00026 51.65 2.16 49.50 79.62 -169 -13.46
Total1945.85 100.43
xux = 425 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 425/450 = 0.340
Puc = 0.340 x 20.00 x 300 x 450
= 918.84 kN
Pux1 = Puc + Pus(Total)
= 918.84 + (1945.85)
= 2864.69 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 425/450
= 0.393
Muc = 918.84 x (0.5 x 450 - 0.393 x 450) = 44.14 kN-m
Mux1 = Muc + Mus(Total)
= 44.14 + (100.43)
= 144.57 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 4021 0.00285 352.56 8.92 343.64 1381.85 94 129.89
2 4021 0.00065 130.67 4.88 125.79 505.83 -94 -47.55
Total1887.69 82.35
xuy = 300 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 300/300 = 0.360
Puc = 0.360 x 20.00 x 450 x 300
= 972.00 kN
Puy1 = Puc + Pus(Total)
= 972.00 + (1887.69)
= 2859.69 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 300/300
= 0.416
Muc = 972.00 x (0.5 x 300 - 0.416 x 300) = 24.49 kN-m
Muy1 = Muc + Mus(Total)
= 24.49 + (82.35)
= 106.84 kN-m
Pu/Puz = 0.780
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.780, an = 1.966
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((71.24/144.57)1.966) + ((56.85/106.84)1.966)
= 0.538 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 32/4 = 8 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (8, 6) = 8 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 32 = 512 mm
               iii) 300 mm
               Required spacing = minimum of (300, 512, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG65 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #32 - 10 nos. (8042 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 3499.16 kN
MomentX,(Mx) = 49.18 kN-m
MomentY,(My) = 2.09 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C730.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 76.107 kN.m
My_MinEccen = 69.983 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 76.107 kN.m
My = max(My,My_MinEccen) + MuaddY = 69.983 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 49.18/3499.16 = 14 mm
Actual eccenY = My / P = 2.09/3499.16 = 1 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(14,22) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(1,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #32 - 12 nos. (9651 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1608 0.00304 354.22 8.92 345.30 555.41 169 93.86
2 1608 0.00252 345.74 8.92 336.82 541.77 101 54.94
3 1608 0.00201 328.20 8.92 319.28 513.56 34 17.36
4 1608 0.00150 294.46 8.36 286.09 460.18 -34 -15.55
5 1608 0.00099 197.80 6.64 191.16 307.47 -101 -31.18
6 1608 0.00048 95.44 3.75 91.69 147.48 -169 -24.92
Total2525.87 94.50
xux = 457 mm               Puc = C1.fck.B.D
ku = 1.016
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.083
C1 = 0.446 x (1 - C3/6) = 0.366
Puc = 0.366 x 20.00 x 300 x 450
= 986.88 kN
Pux1 = Puc + Pus(Total)
= 986.88 + (2525.87)
= 3512.75 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.421
Muc = 986.88 x (0.5 x 450 - 0.421 x 450) = 34.93 kN-m
Mux1 = Muc + Mus(Total)
= 34.93 + (94.50)
= 129.43 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 4825 0.00269 350.12 8.92 341.20 1646.46 94 154.77
2 4825 0.00090 179.01 6.20 172.81 833.90 -94 -78.39
Total2480.36 76.38
xuy = 338 mm               Puc = C1.fck.B.D
ku = 1.125
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.769
C1 = 0.446 x (1 - C3/6) = 0.389
Puc = 0.389 x 20.00 x 450 x 300
= 1049.78 kN
Puy1 = Puc + Pus(Total)
= 1049.78 + (2480.36)
= 3530.14 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.447
Muc = 1049.78 x (0.5 x 300 - 0.447 x 300) = 16.55 kN-m
Muy1 = Muc + Mus(Total)
= 16.55 + (76.38)
= 92.93 kN-m
Pu/Puz = 0.847
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.847, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((76.11/129.43)2.000) + ((69.98/92.93)2.000)
= 0.913 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 32/4 = 8 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (8, 6) = 8 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 32 = 512 mm
               iii) 300 mm
               Required spacing = minimum of (300, 512, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG65 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #32 - 12 nos. (9651 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG66 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 583.77 kN
MomentX,(Mx) = 110.64 kN-m
MomentY,(My) = 37.54 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C740.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 12.084 kN.m
My_MinEccen = 11.675 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 110.637 kN.m
My = max(My,My_MinEccen) + MuaddY = 37.539 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 110.64/583.77 = 190 mm
Actual eccenY = My / P = 37.54/583.77 = 64 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(190,21) = 190 mm
eccenY = max(Actual eccenY,eccenYMin) = max(64,20) = 64 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(190 mm) > 0.05 x 300(15 mm)
and eccenY(64 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 16 nos. + #16 - 4 nos. (2614 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00281 352.22 8.92 343.30 138.05 177 24.43
2 226 0.00222 335.73 8.92 326.81 73.92 136 10.06
3 226 0.00166 308.35 8.66 299.69 67.79 97 6.59
4 226 0.00110 219.06 7.09 211.96 47.94 58 2.80
5 226 0.00053 106.83 4.13 102.71 23.23 19 0.45
6 226 -0.00003 -5.39 0.00 -5.39 -1.22 -19 0.02
7 226 -0.00059 -117.61 0.00 -117.61 -26.60 -58 1.55
8 226 -0.00115 -229.83 0.00 -229.83 -51.99 -97 5.05
9 226 -0.00171 -311.71 0.00 -311.71 -70.51 -136 9.60
10 402 -0.00230 -338.77 0.00 -338.77 -136.23 -177 24.11
Total64.40 84.67
xux = 243 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 243/450 = 0.194
Puc = 0.194 x 20.00 x 300 x 450
= 523.97 kN
Pux1 = Puc + Pus(Total)
= 523.97 + (64.40)
= 588.36 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 243/450
= 0.224
Muc = 523.97 x (0.5 x 450 - 0.224 x 450) = 65.02 kN-m
Mux1 = Muc + Mus(Total)
= 65.02 + (84.67)
= 149.69 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1307 0.00251 345.41 8.92 336.49 439.75 102 44.85
2 1307 -0.00169 -310.48 0.00 -310.48 -405.77 -102 41.39
Total33.98 86.24
xuy = 170 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 170/300 = 0.204
Puc = 0.204 x 20.00 x 450 x 300
= 550.55 kN
Puy1 = Puc + Pus(Total)
= 550.55 + (33.98)
= 584.53 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 170/300
= 0.236
Muc = 550.55 x (0.5 x 300 - 0.236 x 300) = 43.67 kN-m
Muy1 = Muc + Mus(Total)
= 43.67 + (86.24)
= 129.91 kN-m
Pu/Puz = 0.291
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.291, an = 1.152
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((110.64/149.69)1.152) + ((37.54/129.91)1.152)
= 0.945 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG66 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 16 nos. + #16 - 4 nos. (2614 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1181.36 kN
MomentX,(Mx) = 99.35 kN-m
MomentY,(My) = 34.93 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C740.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 24.454 kN.m
My_MinEccen = 23.627 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 99.353 kN.m
My = max(My,My_MinEccen) + MuaddY = 34.929 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 99.35/1181.36 = 84 mm
Actual eccenY = My / P = 34.93/1181.36 = 30 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(84,21) = 84 mm
eccenY = max(Actual eccenY,eccenYMin) = max(30,20) = 30 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(84 mm) > 0.05 x 300(15 mm)
and eccenY(30 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #16 - 14 nos. (2815 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00300 353.87 8.92 344.95 138.71 177 24.55
2 402 0.00238 341.65 8.92 332.73 133.80 118 15.79
3 402 0.00176 314.84 8.79 306.05 123.07 59 7.26
4 402 0.00114 228.42 7.28 221.14 88.93 0 0.00
5 402 0.00052 104.76 4.06 100.70 40.49 -59 -2.39
6 402 -0.00009 -18.90 0.00 -18.90 -7.60 -118 0.90
7 402 -0.00071 -142.55 0.00 -142.55 -57.32 -177 10.15
Total460.08 56.26
xux = 334 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 334/450 = 0.267
Puc = 0.267 x 20.00 x 300 x 450
= 721.41 kN
Pux1 = Puc + Pus(Total)
= 721.41 + (460.08)
= 1181.49 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 334/450
= 0.309
Muc = 721.41 x (0.5 x 450 - 0.309 x 450) = 62.09 kN-m
Mux1 = Muc + Mus(Total)
= 62.09 + (56.26)
= 118.34 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1407 0.00279 352.07 8.92 343.15 482.95 102 49.26
2 1407 -0.00023 -45.19 0.00 -45.19 -63.60 -102 6.49
Total419.35 55.75
xuy = 237 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 237/300 = 0.284
Puc = 0.284 x 20.00 x 450 x 300
= 766.97 kN
Puy1 = Puc + Pus(Total)
= 766.97 + (419.35)
= 1186.32 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 237/300
= 0.328
Muc = 766.97 x (0.5 x 300 - 0.328 x 300) = 39.52 kN-m
Muy1 = Muc + Mus(Total)
= 39.52 + (55.75)
= 95.27 kN-m
Pu/Puz = 0.572
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.572, an = 1.620
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((99.35/118.34)1.620) + ((34.93/95.27)1.620)
= 0.950 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 16 = 256 mm
               iii) 300 mm
               Required spacing = minimum of (300, 256, 300) = 256 mm
# Provide ties               8 @ 250 mm c/c

SUMMARY :

CG66 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #16 - 14 nos. (2815 mm˛)
Provide ties #8 @ 250 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1778.77 kN
MomentX,(Mx) = 93.67 kN-m
MomentY,(My) = 31.87 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C740.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 36.821 kN.m
My_MinEccen = 35.575 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 93.672 kN.m
My = max(My,My_MinEccen) + MuaddY = 35.575 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 93.67/1778.77 = 53 mm
Actual eccenY = My / P = 31.87/1778.77 = 18 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(53,21) = 53 mm
eccenY = max(Actual eccenY,eccenYMin) = max(18,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(53 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #20 - 14 nos. (4398 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 628 0.00305 354.32 8.92 345.40 217.02 175 37.98
2 628 0.00252 345.62 8.92 336.70 211.55 117 24.68
3 628 0.00199 327.43 8.92 318.51 200.13 58 11.67
4 628 0.00146 290.94 8.28 282.66 177.60 0 0.00
5 628 0.00094 187.14 6.39 180.74 113.56 -58 -6.62
6 628 0.00041 81.55 3.27 78.28 49.19 -117 -5.74
7 628 -0.00012 -24.04 0.00 -24.04 -15.11 -175 2.64
Total953.94 64.61
xux = 387 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 387/450 = 0.309
Puc = 0.309 x 20.00 x 300 x 450
= 835.31 kN
Pux1 = Puc + Pus(Total)
= 835.31 + (953.94)
= 1789.26 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 387/450
= 0.358
Muc = 835.31 x (0.5 x 450 - 0.358 x 450) = 53.56 kN-m
Mux1 = Muc + Mus(Total)
= 53.56 + (64.61)
= 118.18 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 2199 0.00287 352.74 8.92 343.82 756.10 100 75.61
2 2199 0.00034 67.23 2.75 64.49 141.81 -100 -14.18
Total897.91 61.43
xuy = 277 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 277/300 = 0.332
Puc = 0.332 x 20.00 x 450 x 300
= 896.06 kN
Puy1 = Puc + Pus(Total)
= 896.06 + (897.91)
= 1793.97 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 277/300
= 0.384
Muc = 896.06 x (0.5 x 300 - 0.384 x 300) = 31.32 kN-m
Muy1 = Muc + Mus(Total)
= 31.32 + (61.43)
= 92.75 kN-m
Pu/Puz = 0.699
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.699, an = 1.832
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((93.67/118.18)1.832) + ((35.58/92.75)1.832)
= 0.826 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 20/4 = 5 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (5, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 20 = 320 mm
               iii) 300 mm
               Required spacing = minimum of (300, 320, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG66 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #20 - 14 nos. (4398 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 2374.24 kN
MomentX,(Mx) = 84.42 kN-m
MomentY,(My) = 27.69 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C740.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 49.147 kN.m
My_MinEccen = 47.485 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 84.425 kN.m
My = max(My,My_MinEccen) + MuaddY = 47.485 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 84.42/2374.24 = 36 mm
Actual eccenY = My / P = 27.69/2374.24 = 12 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(36,21) = 36 mm
eccenY = max(Actual eccenY,eccenYMin) = max(12,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(36 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #25 - 12 nos. (5890 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 982 0.00308 354.56 8.92 345.64 339.33 173 58.53
2 982 0.00252 345.54 8.92 336.62 330.48 104 34.20
3 982 0.00196 326.20 8.92 317.28 311.49 35 10.75
4 982 0.00140 279.92 8.12 271.81 266.85 -35 -9.21
5 982 0.00084 168.23 5.93 162.30 159.34 -104 -16.49
6 982 0.00028 56.53 2.34 54.19 53.20 -173 -9.18
Total1460.68 68.61
xux = 432 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 432/450 = 0.346
Puc = 0.346 x 20.00 x 300 x 450
= 934.03 kN
Pux1 = Puc + Pus(Total)
= 934.03 + (1460.68)
= 2394.71 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 432/450
= 0.400
Muc = 934.03 x (0.5 x 450 - 0.400 x 450) = 42.14 kN-m
Mux1 = Muc + Mus(Total)
= 42.14 + (68.61)
= 110.75 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 2945 0.00285 352.61 8.92 343.69 1012.25 98 98.69
2 2945 0.00067 133.43 4.96 128.47 378.39 -98 -36.89
Total1390.64 61.80
xuy = 307 mm               Puc = C1.fck.B.D
ku = 1.023
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.055
C1 = 0.446 x (1 - C3/6) = 0.368
Puc = 0.368 x 20.00 x 450 x 300
= 992.55 kN
Puy1 = Puc + Pus(Total)
= 992.55 + (1390.64)
= 2383.19 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.424
Muc = 992.55 x (0.5 x 300 - 0.424 x 300) = 22.68 kN-m
Muy1 = Muc + Mus(Total)
= 22.68 + (61.80)
= 84.48 kN-m
Pu/Puz = 0.793
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.793, an = 1.988
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((84.42/110.75)1.988) + ((47.48/84.48)1.988)
= 0.901 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 25/4 = 6 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (6, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 25 = 400 mm
               iii) 300 mm
               Required spacing = minimum of (300, 400, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG66 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #25 - 12 nos. (5890 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 2964.67 kN
MomentX,(Mx) = 57.74 kN-m
MomentY,(My) = 15.36 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C740.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 64.482 kN.m
My_MinEccen = 59.293 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 64.482 kN.m
My = max(My,My_MinEccen) + MuaddY = 59.293 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 57.74/2964.67 = 19 mm
Actual eccenY = My / P = 15.36/2964.67 = 5 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(19,22) = 22 mm
eccenY = max(Actual eccenY,eccenYMin) = max(5,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(22 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #32 - 10 nos. (8042 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1608 0.00306 354.40 8.92 345.48 555.71 169 93.91
2 1608 0.00239 342.06 8.92 333.14 535.86 85 45.28
3 1608 0.00172 312.46 8.75 303.71 488.51 0 0.00
4 1608 0.00105 210.91 6.93 203.99 328.11 -85 -27.73
5 1608 0.00039 77.38 3.12 74.27 119.46 -169 -20.19
Total2027.64 91.28
xux = 443 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 443/450 = 0.354
Puc = 0.354 x 20.00 x 300 x 450
= 956.81 kN
Pux1 = Puc + Pus(Total)
= 956.81 + (2027.64)
= 2984.46 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 443/450
= 0.410
Muc = 956.81 x (0.5 x 450 - 0.410 x 450) = 38.97 kN-m
Mux1 = Muc + Mus(Total)
= 38.97 + (91.28)
= 130.25 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 4021 0.00278 352.00 8.92 343.08 1379.59 94 129.68
2 4021 0.00076 151.09 5.47 145.62 585.57 -94 -55.04
Total1965.16 74.64
xuy = 314 mm               Puc = C1.fck.B.D
ku = 1.047
C3 = (8/7) x [4/(7.ku - 3)]2 = 0.976
C1 = 0.446 x (1 - C3/6) = 0.373
Puc = 0.373 x 20.00 x 450 x 300
= 1008.29 kN
Puy1 = Puc + Pus(Total)
= 1008.29 + (1965.16)
= 2973.45 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.431
Muc = 1008.29 x (0.5 x 300 - 0.431 x 300) = 20.99 kN-m
Muy1 = Muc + Mus(Total)
= 20.99 + (74.64)
= 95.63 kN-m
Pu/Puz = 0.813
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.813, an = 2.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((64.48/130.25)2.000) + ((59.29/95.63)2.000)
= 0.630 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 32/4 = 8 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (8, 6) = 8 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 32 = 512 mm
               iii) 300 mm
               Required spacing = minimum of (300, 512, 300) = 300 mm
# Provide ties               8 @ 300 mm c/c

SUMMARY :

CG66 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #32 - 10 nos. (8042 mm˛)
Provide ties #8 @ 300 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG67 :

General Design Parameters :
Below Floor5 - Default Level at 19.500 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 245.67 kN
MomentX,(Mx) = 79.05 kN-m
MomentY,(My) = 41.47 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C750.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 5.085 kN.m
My_MinEccen = 4.913 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 79.055 kN.m
My = max(My,My_MinEccen) + MuaddY = 41.472 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 79.05/245.67 = 322 mm
Actual eccenY = My / P = 41.47/245.67 = 169 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(322,21) = 322 mm
eccenY = max(Actual eccenY,eccenYMin) = max(169,20) = 169 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(322 mm) > 0.05 x 300(15 mm)
and eccenY(169 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 4 nos. + #16 - 8 nos. (2061 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 402 0.00256 346.73 8.92 337.81 135.84 177 24.04
2 402 0.00115 229.94 7.31 222.63 89.53 105 9.36
3 226 -0.00022 -44.91 0.00 -44.91 -10.16 34 -0.35
4 226 -0.00156 -300.05 0.00 -300.05 -67.87 -34 2.32
5 402 -0.00293 -353.32 0.00 -353.32 -142.08 -105 14.86
6 402 -0.00435 -360.90 0.00 -360.90 -145.13 -177 25.69
Total-139.86 75.93
xux = 179 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 179/450 = 0.143
Puc = 0.143 x 20.00 x 300 x 450
= 387.28 kN
Pux1 = Puc + Pus(Total)
= 387.28 + (-139.86)
= 247.42 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 179/450
= 0.166
Muc = 387.28 x (0.5 x 450 - 0.166 x 450) = 58.25 kN-m
Mux1 = Muc + Mus(Total)
= 58.25 + (75.93)
= 134.18 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 1030 0.00172 312.26 8.74 303.52 312.76 102 31.90
2 1030 -0.00585 -360.90 0.00 -360.90 -371.89 -102 37.93
Total-59.13 69.83
xuy = 94 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 94/300 = 0.113
Puc = 0.113 x 20.00 x 450 x 300
= 305.65 kN
Puy1 = Puc + Pus(Total)
= 305.65 + (-59.13)
= 246.52 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 94/300
= 0.131
Muc = 305.65 x (0.5 x 300 - 0.131 x 300) = 33.85 kN-m
Muy1 = Muc + Mus(Total)
= 33.85 + (69.83)
= 103.69 kN-m
Pu/Puz = 0.134
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.134, an = 1.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((79.05/134.18)1.000) + ((41.47/103.69)1.000)
= 0.989 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 16/4 = 4 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (4, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG67 : Floor5 - Default Level at 19.500 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 4 nos. + #16 - 8 nos. (2061 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor4 - Default Level at 15.600 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 519.73 kN
MomentX,(Mx) = 69.42 kN-m
MomentY,(My) = 37.19 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C750.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 10.758 kN.m
My_MinEccen = 10.395 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 69.424 kN.m
My = max(My,My_MinEccen) + MuaddY = 37.192 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 69.42/519.73 = 134 mm
Actual eccenY = My / P = 37.19/519.73 = 72 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(134,21) = 134 mm
eccenY = max(Actual eccenY,eccenYMin) = max(72,20) = 72 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(134 mm) > 0.05 x 300(15 mm)
and eccenY(72 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00281 352.27 8.92 343.35 77.66 179 13.90
2 226 0.00148 292.40 8.31 284.09 64.26 90 5.75
3 226 0.00014 28.84 1.24 27.60 6.24 0 0.00
4 226 -0.00119 -238.13 0.00 -238.13 -53.86 -90 4.82
5 226 -0.00253 -345.77 0.00 -345.77 -78.21 -179 14.00
Total16.09 38.47
xux = 235 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 235/450 = 0.188
Puc = 0.188 x 20.00 x 300 x 450
= 506.88 kN
Pux1 = Puc + Pus(Total)
= 506.88 + (16.09)
= 522.97 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 235/450
= 0.217
Muc = 506.88 x (0.5 x 450 - 0.217 x 450) = 64.57 kN-m
Mux1 = Muc + Mus(Total)
= 64.57 + (38.47)
= 103.04 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00250 345.04 8.92 336.12 190.07 104 19.77
2 565 -0.00204 -329.11 0.00 -329.11 -186.11 -104 19.36
Total3.97 39.12
xuy = 161 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 161/300 = 0.193
Puc = 0.193 x 20.00 x 450 x 300
= 520.17 kN
Puy1 = Puc + Pus(Total)
= 520.17 + (3.97)
= 524.14 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 161/300
= 0.223
Muc = 520.17 x (0.5 x 300 - 0.223 x 300) = 43.28 kN-m
Muy1 = Muc + Mus(Total)
= 43.28 + (39.12)
= 82.41 kN-m
Pu/Puz = 0.334
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.334, an = 1.223
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((69.42/103.04)1.223) + ((37.19/82.41)1.223)
= 0.995 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG67 : Floor4 - Default Level at 15.600 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor3 - Default Level at 11.700 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 786.83 kN
MomentX,(Mx) = 66.20 kN-m
MomentY,(My) = 32.50 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C750.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 16.287 kN.m
My_MinEccen = 15.737 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 66.199 kN.m
My = max(My,My_MinEccen) + MuaddY = 32.498 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 66.20/786.83 = 84 mm
Actual eccenY = My / P = 32.50/786.83 = 41 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(84,21) = 84 mm
eccenY = max(Actual eccenY,eccenYMin) = max(41,20) = 41 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(84 mm) > 0.05 x 300(15 mm)
and eccenY(41 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00297 353.64 8.92 344.72 77.97 179 13.96
2 226 0.00194 325.55 8.91 316.64 71.62 90 6.41
3 226 0.00091 182.08 6.27 175.81 39.77 0 0.00
4 226 -0.00012 -23.94 0.00 -23.94 -5.41 -90 0.48
5 226 -0.00115 -229.95 0.00 -229.95 -52.01 -179 9.31
Total131.94 30.16
xux = 304 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 304/450 = 0.243
Puc = 0.243 x 20.00 x 300 x 450
= 656.86 kN
Pux1 = Puc + Pus(Total)
= 656.86 + (131.94)
= 788.79 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 304/450
= 0.281
Muc = 656.86 x (0.5 x 450 - 0.281 x 450) = 64.70 kN-m
Mux1 = Muc + Mus(Total)
= 64.70 + (30.16)
= 94.86 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00273 351.09 8.92 342.17 193.49 104 20.12
2 565 -0.00074 -147.61 0.00 -147.61 -83.47 -104 8.68
Total110.02 28.80
xuy = 210 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 210/300 = 0.252
Puc = 0.252 x 20.00 x 450 x 300
= 679.64 kN
Puy1 = Puc + Pus(Total)
= 679.64 + (110.02)
= 789.66 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 210/300
= 0.291
Muc = 679.64 x (0.5 x 300 - 0.291 x 300) = 42.64 kN-m
Muy1 = Muc + Mus(Total)
= 42.64 + (28.80)
= 71.44 kN-m
Pu/Puz = 0.505
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.505, an = 1.509
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((66.20/94.86)1.509) + ((32.50/71.44)1.509)
= 0.886 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG67 : Floor3 - Default Level at 11.700 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 2850 mm
From analysis results,loads on column
Axial load,(P) = 1044.08 kN
MomentX,(Mx) = 61.24 kN-m
MomentY,(My) = 25.87 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C750.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 2850 x 1.000 ) / 450 = 6.33 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 2850 x 1.000 ) / 300 = 9.50 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 21.613 kN.m
My_MinEccen = 20.882 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 61.241 kN.m
My = max(My,My_MinEccen) + MuaddY = 25.869 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 61.24/1044.08 = 59 mm
Actual eccenY = My / P = 25.87/1044.08 = 25 mm
eccenXMin = (L/500) + (D/30) = (2850/500) + (450/30) = 21 mm
eccenXMin = max(21,20)= 21 mm
eccenYMin = (L/500) + (B/30) = (2850/500) + (300/30) = 16 mm
eccenYMin = max(16,20) = 21 mm
eccenX = max(Actual eccenX,eccenXMin) = max(59,21) = 59 mm
eccenY = max(Actual eccenY,eccenYMin) = max(25,20) = 25 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(59 mm) > 0.05 x 300(15 mm)
and eccenY(25 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00308 354.56 8.92 345.64 78.18 179 13.99
2 226 0.00225 336.96 8.92 328.04 74.20 90 6.64
3 226 0.00143 285.19 8.19 277.00 62.66 0 0.00
4 226 0.00060 120.18 4.55 115.63 26.15 -90 -2.34
5 226 -0.00022 -44.82 0.00 -44.82 -10.14 -179 1.81
Total231.05 20.11
xux = 380 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 380/450 = 0.304
Puc = 0.304 x 20.00 x 300 x 450
= 820.13 kN
Pux1 = Puc + Pus(Total)
= 820.13 + (231.05)
= 1051.18 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 380/450
= 0.351
Muc = 820.13 x (0.5 x 450 - 0.351 x 450) = 54.99 kN-m
Mux1 = Muc + Mus(Total)
= 54.99 + (20.11)
= 75.10 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00288 352.86 8.92 343.94 194.49 104 20.23
2 565 0.00008 16.56 0.72 15.84 8.96 -104 -0.93
Total203.45 19.30
xuy = 260 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 260/300 = 0.312
Puc = 0.312 x 20.00 x 450 x 300
= 842.91 kN
Puy1 = Puc + Pus(Total)
= 842.91 + (203.45)
= 1046.36 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 260/300
= 0.361
Muc = 842.91 x (0.5 x 300 - 0.361 x 300) = 35.21 kN-m
Muy1 = Muc + Mus(Total)
= 35.21 + (19.30)
= 54.51 kN-m
Pu/Puz = 0.671
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.671, an = 1.784
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((61.24/75.10)1.784) + ((25.87/54.51)1.784)
= 0.959 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG67 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.



General Design Parameters :
Below Floor1 - Default Level at 3.900 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 3375 mm
From analysis results,loads on column
Axial load,(P) = 1284.26 kN
MomentX,(Mx) = 41.39 kN-m
MomentY,(My) = 12.86 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C750.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 3375 x 1.000 ) / 450 = 7.50 <= 12.00,
column is not slender in this direction.
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 3375 x 1.000 ) / 300 = 11.25 <= 12.00,
column is not slender in this direction.
Mx_MinEccen = 27.933 kN.m
My_MinEccen = 25.685 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 41.394 kN.m
My = max(My,My_MinEccen) + MuaddY = 25.685 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 41.39/1284.26 = 32 mm
Actual eccenY = My / P = 12.86/1284.26 = 10 mm
eccenXMin = (L/500) + (D/30) = (3375/500) + (450/30) = 22 mm
eccenXMin = max(22,20)= 22 mm
eccenYMin = (L/500) + (B/30) = (3375/500) + (300/30) = 17 mm
eccenYMin = max(17,20) = 22 mm
eccenX = max(Actual eccenX,eccenXMin) = max(32,22) = 32 mm
eccenY = max(Actual eccenY,eccenYMin) = max(10,20) = 20 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(32 mm) > 0.05 x 300(15 mm)
and eccenY(20 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 12 nos. (1357 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00313 355.07 8.92 346.15 78.30 179 14.02
2 226 0.00256 346.74 8.92 337.82 76.41 107 8.21
3 226 0.00199 327.49 8.92 318.57 72.06 36 2.58
4 226 0.00142 284.57 8.18 276.40 62.52 -36 -2.24
5 226 0.00085 170.52 5.98 164.54 37.22 -107 -4.00
6 226 0.00028 56.47 2.34 54.13 12.24 -179 -2.19
Total338.75 16.37
xux = 439 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 439/450 = 0.352
Puc = 0.352 x 20.00 x 300 x 450
= 949.22 kN
Pux1 = Puc + Pus(Total)
= 949.22 + (338.75)
= 1287.97 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 439/450
= 0.406
Muc = 949.22 x (0.5 x 450 - 0.406 x 450) = 40.05 kN-m
Mux1 = Muc + Mus(Total)
= 40.05 + (16.37)
= 56.42 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 679 0.00294 353.35 8.92 344.43 233.73 104 24.31
2 679 0.00058 115.12 4.40 110.73 75.14 -104 -7.81
Total308.87 16.49
xuy = 305 mm               Puc = C1.fck.B.D
ku = 1.016
C3 = (8/7) x [4/(7.ku - 3)]2 = 1.083
C1 = 0.446 x (1 - C3/6) = 0.366
Puc = 0.366 x 20.00 x 450 x 300
= 986.88 kN
Puy1 = Puc + Pus(Total)
= 986.88 + (308.87)
= 1295.74 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = (0.5f - C3/7)/(1 - C3/6) = 0.421
Muc = 986.88 x (0.5 x 300 - 0.421 x 300) = 23.28 kN-m
Muy1 = Muc + Mus(Total)
= 23.28 + (16.49)
= 39.78 kN-m
Pu/Puz = 0.790
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.790, an = 1.984
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((41.39/56.42)1.984) + ((25.69/39.78)1.984)
= 0.961 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG67 : Floor1 - Default Level at 3.900 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 12 nos. (1357 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG68 :

General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 7275 mm
From analysis results,loads on column
Axial load,(P) = 99.80 kN
MomentX,(Mx) = 13.02 kN-m
MomentY,(My) = 1.43 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C30.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 7275 x 1.000 ) / 450 = 16.17 > 12.00,
column is slender in this direction.
Puz = (0.45 x fck x Ag) + (0.75 x fy - 0.45 x fck) x Ast
= 0.45 x 20.00 x 135000.01 + (0.75 x 415.00 - 0.45 x 20.00 ) x 1130.97 = 1556836.792 N
k1 = 0.206511
k2 = 0.080400 MPa
Pub = (k1 x fck x Ag) + (k2 x Ast x 100)
= (0.206511 x 20.00 x 135000.01) + (0.08 x 1130.97 x 100) = 566673.035 N
k = (Puz - Pu) / (Puz - Pub) <= 1
= (1556836.792 - 99803.696) / (1556836.792 - 566673.035) = 1.471507
k = 1.471507 > 1 hence k = 1
MuaddX = Pu x (Depth/2000) x (Lo/Depth) x (Lo/Depth) x k
= 99803.70 x (450.00/2000) x (7275.00/450.00) x (7275.00/450.00) x 1.000000
= 5869081.497 N.mm = 5.869 kN.m
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 7275 x 1.000 ) / 300 = 24.25 > 12.00,
column is slender in this direction.
Puz = (0.45 x fck x Ag) + (0.75 x fy - 0.45 x fck) x Ast
= 0.45 x 20.00 x 135000.01 + (0.75 x 415.00 - 0.45 x 20.00 ) x 1130.97 = 1556836.792 N
k1 = 0.195200
k2 = 0.041467 MPa
Pub = (k1 x fck x Ag) + (k2 x Ast x 100)
= (0.195200 x 20.00 x 135000.01) + (0.04 x 1130.97 x 100) = 531729.797 N
k = (Puz - Pu) / (Puz - Pub) <= 1
= (1556836.792 - 99803.696) / (1556836.792 - 531729.797) = 1.421347
k = 1.421347 > 1 hence k = 1
MuaddY = Pu x (Width/2000) x (Lo/Width) x (Lo/Width) x k
= 99803.70 x (300.00/2000) x (7275.00/300.00) x (7275.00/300.00) x 1.000000
= 8803621.292 N.mm = 8.804 kN.m
Mx_MinEccen = 2.949 kN.m
My_MinEccen = 2.450 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 18.890 kN.m
My = max(My,My_MinEccen) + MuaddY = 11.254 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 13.02/99.80 = 130 mm
Actual eccenY = My / P = 1.43/99.80 = 14 mm
eccenXMin = (L/500) + (D/30) = (7275/500) + (450/30) = 30 mm
eccenXMin = max(30,20)= 30 mm
eccenYMin = (L/500) + (B/30) = (7275/500) + (300/30) = 25 mm
eccenYMin = max(25,20) = 30 mm
eccenX = max(Actual eccenX,eccenXMin) = max(130,30) = 130 mm
eccenY = max(Actual eccenY,eccenYMin) = max(14,25) = 25 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(130 mm) > 0.05 x 300(15 mm)
and eccenY(25 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00225 336.75 8.92 327.83 74.15 179 13.27
2 226 -0.00020 -39.17 0.00 -39.17 -8.86 90 -0.79
3 226 -0.00264 -348.64 0.00 -348.64 -78.86 0 0.00
4 226 -0.00508 -360.90 0.00 -360.90 -81.63 -90 7.31
5 226 -0.00752 -360.90 0.00 -360.90 -81.63 -179 14.61
Total-176.83 34.40
xux = 128 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 128/450 = 0.103
Puc = 0.103 x 20.00 x 300 x 450
= 277.17 kN
Pux1 = Puc + Pus(Total)
= 277.17 + (-176.83)
= 100.34 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 128/450
= 0.119
Muc = 277.17 x (0.5 x 450 - 0.119 x 450) = 47.57 kN-m
Mux1 = Muc + Mus(Total)
= 47.57 + (34.40)
= 81.97 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00093 185.80 6.36 179.44 101.47 104 10.55
2 565 -0.01070 -360.90 0.00 -360.90 -204.08 -104 21.22
Total-102.61 31.78
xuy = 63 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 63/300 = 0.075
Puc = 0.075 x 20.00 x 450 x 300
= 202.90 kN
Puy1 = Puc + Pus(Total)
= 202.90 + (-102.61)
= 100.28 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 63/300
= 0.087
Muc = 202.90 x (0.5 x 300 - 0.087 x 300) = 25.15 kN-m
Muy1 = Muc + Mus(Total)
= 25.15 + (31.78)
= 56.93 kN-m
Pu/Puz = 0.064
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.064, an = 1.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((18.89/81.97)1.000) + ((11.25/56.93)1.000)
= 0.428 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG68 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG69 :

General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 7275 mm
From analysis results,loads on column
Axial load,(P) = 317.54 kN
MomentX,(Mx) = 21.10 kN-m
MomentY,(My) = 4.19 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C50.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 7275 x 1.000 ) / 450 = 16.17 > 12.00,
column is slender in this direction.
Puz = (0.45 x fck x Ag) + (0.75 x fy - 0.45 x fck) x Ast
= 0.45 x 20.00 x 135000.01 + (0.75 x 415.00 - 0.45 x 20.00 ) x 1130.97 = 1556836.792 N
k1 = 0.206511
k2 = 0.080400 MPa
Pub = (k1 x fck x Ag) + (k2 x Ast x 100)
= (0.206511 x 20.00 x 135000.01) + (0.08 x 1130.97 x 100) = 566673.035 N
k = (Puz - Pu) / (Puz - Pub) <= 1
= (1556836.792 - 317535.889) / (1556836.792 - 566673.035) = 1.251612
k = 1.251612 > 1 hence k = 1
MuaddX = Pu x (Depth/2000) x (Lo/Depth) x (Lo/Depth) x k
= 317535.89 x (450.00/2000) x (7275.00/450.00) x (7275.00/450.00) x 1.000000
= 18673095.703 N.mm = 18.673 kN.m
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 7275 x 1.000 ) / 300 = 24.25 > 12.00,
column is slender in this direction.
Puz = (0.45 x fck x Ag) + (0.75 x fy - 0.45 x fck) x Ast
= 0.45 x 20.00 x 135000.01 + (0.75 x 415.00 - 0.45 x 20.00 ) x 1130.97 = 1556836.792 N
k1 = 0.195200
k2 = 0.041467 MPa
Pub = (k1 x fck x Ag) + (k2 x Ast x 100)
= (0.195200 x 20.00 x 135000.01) + (0.04 x 1130.97 x 100) = 531729.797 N
k = (Puz - Pu) / (Puz - Pub) <= 1
= (1556836.792 - 317535.889) / (1556836.792 - 531729.797) = 1.208948
k = 1.208948 > 1 hence k = 1
MuaddY = Pu x (Width/2000) x (Lo/Width) x (Lo/Width) x k
= 317535.89 x (300.00/2000) x (7275.00/300.00) x (7275.00/300.00) x 1.000000
= 28009641.647 N.mm = 28.010 kN.m
Mx_MinEccen = 9.383 kN.m
My_MinEccen = 7.796 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 39.777 kN.m
My = max(My,My_MinEccen) + MuaddY = 35.805 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 21.10/317.54 = 66 mm
Actual eccenY = My / P = 4.19/317.54 = 13 mm
eccenXMin = (L/500) + (D/30) = (7275/500) + (450/30) = 30 mm
eccenXMin = max(30,20)= 30 mm
eccenYMin = (L/500) + (B/30) = (7275/500) + (300/30) = 25 mm
eccenYMin = max(25,20) = 30 mm
eccenX = max(Actual eccenX,eccenXMin) = max(66,30) = 66 mm
eccenY = max(Actual eccenY,eccenYMin) = max(13,25) = 25 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(66 mm) > 0.05 x 300(15 mm)
and eccenY(25 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00262 348.29 8.92 339.37 76.76 179 13.74
2 226 0.00092 183.64 6.31 177.33 40.11 90 3.59
3 226 -0.00079 -157.42 0.00 -157.42 -35.61 0 0.00
4 226 -0.00249 -344.92 0.00 -344.92 -78.02 -90 6.98
5 226 -0.00420 -360.90 0.00 -360.90 -81.63 -179 14.61
Total-78.38 38.93
xux = 184 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 184/450 = 0.147
Puc = 0.147 x 20.00 x 300 x 450
= 396.77 kN
Pux1 = Puc + Pus(Total)
= 396.77 + (-78.38)
= 318.39 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 184/450
= 0.170
Muc = 396.77 x (0.5 x 450 - 0.170 x 450) = 58.95 kN-m
Mux1 = Muc + Mus(Total)
= 58.95 + (38.93)
= 97.88 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00198 327.07 8.92 318.16 179.91 104 18.71
2 565 -0.00488 -360.90 0.00 -360.90 -204.08 -104 21.22
Total-24.17 39.94
xuy = 106 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 106/300 = 0.127
Puc = 0.127 x 20.00 x 450 x 300
= 343.62 kN
Puy1 = Puc + Pus(Total)
= 343.62 + (-24.17)
= 319.45 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 106/300
= 0.147
Muc = 343.62 x (0.5 x 300 - 0.147 x 300) = 36.38 kN-m
Muy1 = Muc + Mus(Total)
= 36.38 + (39.94)
= 76.32 kN-m
Pu/Puz = 0.204
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.204, an = 1.007
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((39.78/97.88)1.007) + ((35.81/76.32)1.007)
= 0.871 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG69 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG70 :

General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 7275 mm
From analysis results,loads on column
Axial load,(P) = 311.80 kN
MomentX,(Mx) = 18.18 kN-m
MomentY,(My) = 3.44 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C60.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 7275 x 1.000 ) / 450 = 16.17 > 12.00,
column is slender in this direction.
Puz = (0.45 x fck x Ag) + (0.75 x fy - 0.45 x fck) x Ast
= 0.45 x 20.00 x 135000.01 + (0.75 x 415.00 - 0.45 x 20.00 ) x 1130.97 = 1556836.792 N
k1 = 0.206511
k2 = 0.080400 MPa
Pub = (k1 x fck x Ag) + (k2 x Ast x 100)
= (0.206511 x 20.00 x 135000.01) + (0.08 x 1130.97 x 100) = 566673.035 N
k = (Puz - Pu) / (Puz - Pub) <= 1
= (1556836.792 - 311804.077) / (1556836.792 - 566673.035) = 1.257401
k = 1.257401 > 1 hence k = 1
MuaddX = Pu x (Depth/2000) x (Lo/Depth) x (Lo/Depth) x k
= 311804.08 x (450.00/2000) x (7275.00/450.00) x (7275.00/450.00) x 1.000000
= 18336029.053 N.mm = 18.336 kN.m
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 7275 x 1.000 ) / 300 = 24.25 > 12.00,
column is slender in this direction.
Puz = (0.45 x fck x Ag) + (0.75 x fy - 0.45 x fck) x Ast
= 0.45 x 20.00 x 135000.01 + (0.75 x 415.00 - 0.45 x 20.00 ) x 1130.97 = 1556836.792 N
k1 = 0.195200
k2 = 0.041467 MPa
Pub = (k1 x fck x Ag) + (k2 x Ast x 100)
= (0.195200 x 20.00 x 135000.01) + (0.04 x 1130.97 x 100) = 531729.797 N
k = (Puz - Pu) / (Puz - Pub) <= 1
= (1556836.792 - 311804.077) / (1556836.792 - 531729.797) = 1.214539
k = 1.214539 > 1 hence k = 1
MuaddY = Pu x (Width/2000) x (Lo/Width) x (Lo/Width) x k
= 311804.08 x (300.00/2000) x (7275.00/300.00) x (7275.00/300.00) x 1.000000
= 27504041.672 N.mm = 27.504 kN.m
Mx_MinEccen = 9.214 kN.m
My_MinEccen = 7.655 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 36.516 kN.m
My = max(My,My_MinEccen) + MuaddY = 35.159 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 18.18/311.80 = 58 mm
Actual eccenY = My / P = 3.44/311.80 = 11 mm
eccenXMin = (L/500) + (D/30) = (7275/500) + (450/30) = 30 mm
eccenXMin = max(30,20)= 30 mm
eccenYMin = (L/500) + (B/30) = (7275/500) + (300/30) = 25 mm
eccenYMin = max(25,20) = 30 mm
eccenX = max(Actual eccenX,eccenXMin) = max(58,30) = 58 mm
eccenY = max(Actual eccenY,eccenYMin) = max(11,25) = 25 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(58 mm) > 0.05 x 300(15 mm)
and eccenY(25 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00262 348.18 8.92 339.26 76.74 179 13.74
2 226 0.00091 181.16 6.25 174.91 39.56 90 3.54
3 226 -0.00081 -161.54 0.00 -161.54 -36.54 0 0.00
4 226 -0.00252 -345.66 0.00 -345.66 -78.19 -90 7.00
5 226 -0.00423 -360.90 0.00 -360.90 -81.63 -179 14.61
Total-80.06 38.89
xux = 183 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 183/450 = 0.146
Puc = 0.146 x 20.00 x 300 x 450
= 394.88 kN
Pux1 = Puc + Pus(Total)
= 394.88 + (-80.06)
= 314.82 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 183/450
= 0.169
Muc = 394.88 x (0.5 x 450 - 0.169 x 450) = 58.82 kN-m
Mux1 = Muc + Mus(Total)
= 58.82 + (38.89)
= 97.70 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00196 326.13 8.92 317.22 179.38 104 18.66
2 565 -0.00502 -360.90 0.00 -360.90 -204.08 -104 21.22
Total-24.70 39.88
xuy = 104 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 104/300 = 0.125
Puc = 0.125 x 20.00 x 450 x 300
= 337.92 kN
Puy1 = Puc + Pus(Total)
= 337.92 + (-24.70)
= 313.22 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 104/300
= 0.145
Muc = 337.92 x (0.5 x 300 - 0.145 x 300) = 36.03 kN-m
Muy1 = Muc + Mus(Total)
= 36.03 + (39.88)
= 75.91 kN-m
Pu/Puz = 0.200
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.200, an = 1.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((36.52/97.70)1.000) + ((35.16/75.91)1.000)
= 0.837 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG70 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG71 :

General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 7275 mm
From analysis results,loads on column
Axial load,(P) = 115.81 kN
MomentX,(Mx) = 25.12 kN-m
MomentY,(My) = 0.31 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C80.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 7275 x 1.000 ) / 450 = 16.17 > 12.00,
column is slender in this direction.
Puz = (0.45 x fck x Ag) + (0.75 x fy - 0.45 x fck) x Ast
= 0.45 x 20.00 x 135000.01 + (0.75 x 415.00 - 0.45 x 20.00 ) x 1130.97 = 1556836.792 N
k1 = 0.206511
k2 = 0.080400 MPa
Pub = (k1 x fck x Ag) + (k2 x Ast x 100)
= (0.206511 x 20.00 x 135000.01) + (0.08 x 1130.97 x 100) = 566673.035 N
k = (Puz - Pu) / (Puz - Pub) <= 1
= (1556836.792 - 115810.799) / (1556836.792 - 566673.035) = 1.455341
k = 1.455341 > 1 hence k = 1
MuaddX = Pu x (Depth/2000) x (Lo/Depth) x (Lo/Depth) x k
= 115810.80 x (450.00/2000) x (7275.00/450.00) x (7275.00/450.00) x 1.000000
= 6810399.055 N.mm = 6.810 kN.m
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 7275 x 1.000 ) / 300 = 24.25 > 12.00,
column is slender in this direction.
Puz = (0.45 x fck x Ag) + (0.75 x fy - 0.45 x fck) x Ast
= 0.45 x 20.00 x 135000.01 + (0.75 x 415.00 - 0.45 x 20.00 ) x 1130.97 = 1556836.792 N
k1 = 0.195200
k2 = 0.041467 MPa
Pub = (k1 x fck x Ag) + (k2 x Ast x 100)
= (0.195200 x 20.00 x 135000.01) + (0.04 x 1130.97 x 100) = 531729.797 N
k = (Puz - Pu) / (Puz - Pub) <= 1
= (1556836.792 - 115810.799) / (1556836.792 - 531729.797) = 1.405732
k = 1.405732 > 1 hence k = 1
MuaddY = Pu x (Width/2000) x (Lo/Width) x (Lo/Width) x k
= 115810.80 x (300.00/2000) x (7275.00/300.00) x (7275.00/300.00) x 1.000000
= 10215598.106 N.mm = 10.216 kN.m
Mx_MinEccen = 3.422 kN.m
My_MinEccen = 2.843 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 31.933 kN.m
My = max(My,My_MinEccen) + MuaddY = 13.059 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 25.12/115.81 = 217 mm
Actual eccenY = My / P = 0.31/115.81 = 3 mm
eccenXMin = (L/500) + (D/30) = (7275/500) + (450/30) = 30 mm
eccenXMin = max(30,20)= 30 mm
eccenYMin = (L/500) + (B/30) = (7275/500) + (300/30) = 25 mm
eccenYMin = max(25,20) = 30 mm
eccenX = max(Actual eccenX,eccenXMin) = max(217,30) = 217 mm
eccenY = max(Actual eccenY,eccenYMin) = max(3,25) = 25 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(217 mm) > 0.05 x 300(15 mm)
and eccenY(25 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00229 338.28 8.92 329.36 74.50 179 13.34
2 226 -0.00007 -14.69 0.00 -14.69 -3.32 90 -0.30
3 226 -0.00243 -343.41 0.00 -343.41 -77.68 0 0.00
4 226 -0.00479 -360.90 0.00 -360.90 -81.63 -90 7.31
5 226 -0.00715 -360.90 0.00 -360.90 -81.63 -179 14.61
Total-169.77 34.96
xux = 133 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 133/450 = 0.106
Puc = 0.106 x 20.00 x 300 x 450
= 286.66 kN
Pux1 = Puc + Pus(Total)
= 286.66 + (-169.77)
= 116.89 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 133/450
= 0.123
Muc = 286.66 x (0.5 x 450 - 0.123 x 450) = 48.67 kN-m
Mux1 = Muc + Mus(Total)
= 48.67 + (34.96)
= 83.63 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00101 202.67 6.75 195.92 110.79 104 11.52
2 565 -0.01023 -360.90 0.00 -360.90 -204.08 -104 21.22
Total-93.29 32.75
xuy = 65 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 65/300 = 0.078
Puc = 0.078 x 20.00 x 450 x 300
= 209.78 kN
Puy1 = Puc + Pus(Total)
= 209.78 + (-93.29)
= 116.49 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 65/300
= 0.090
Muc = 209.78 x (0.5 x 300 - 0.090 x 300) = 25.82 kN-m
Muy1 = Muc + Mus(Total)
= 25.82 + (32.75)
= 58.56 kN-m
Pu/Puz = 0.074
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.074, an = 1.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((31.93/83.63)1.000) + ((13.06/58.56)1.000)
= 0.605 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG71 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG72 :

General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 7275 mm
From analysis results,loads on column
Axial load,(P) = 221.55 kN
MomentX,(Mx) = 1.27 kN-m
MomentY,(My) = 2.54 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C40.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 7275 x 1.000 ) / 450 = 16.17 > 12.00,
column is slender in this direction.
Puz = (0.45 x fck x Ag) + (0.75 x fy - 0.45 x fck) x Ast
= 0.45 x 20.00 x 135000.01 + (0.75 x 415.00 - 0.45 x 20.00 ) x 1130.97 = 1556836.792 N
k1 = 0.206511
k2 = 0.080400 MPa
Pub = (k1 x fck x Ag) + (k2 x Ast x 100)
= (0.206511 x 20.00 x 135000.01) + (0.08 x 1130.97 x 100) = 566673.035 N
k = (Puz - Pu) / (Puz - Pub) <= 1
= (1556836.792 - 221551.804) / (1556836.792 - 566673.035) = 1.348550
k = 1.348550 > 1 hence k = 1
MuaddX = Pu x (Depth/2000) x (Lo/Depth) x (Lo/Depth) x k
= 221551.80 x (450.00/2000) x (7275.00/450.00) x (7275.00/450.00) x 1.000000
= 13028631.210 N.mm = 13.029 kN.m
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 7275 x 1.000 ) / 300 = 24.25 > 12.00,
column is slender in this direction.
Puz = (0.45 x fck x Ag) + (0.75 x fy - 0.45 x fck) x Ast
= 0.45 x 20.00 x 135000.01 + (0.75 x 415.00 - 0.45 x 20.00 ) x 1130.97 = 1556836.792 N
k1 = 0.195200
k2 = 0.041467 MPa
Pub = (k1 x fck x Ag) + (k2 x Ast x 100)
= (0.195200 x 20.00 x 135000.01) + (0.04 x 1130.97 x 100) = 531729.797 N
k = (Puz - Pu) / (Puz - Pub) <= 1
= (1556836.792 - 221551.804) / (1556836.792 - 531729.797) = 1.302581
k = 1.302581 > 1 hence k = 1
MuaddY = Pu x (Width/2000) x (Lo/Width) x (Lo/Width) x k
= 221551.80 x (300.00/2000) x (7275.00/300.00) x (7275.00/300.00) x 1.000000
= 19542945.862 N.mm = 19.543 kN.m
Mx_MinEccen = 6.547 kN.m
My_MinEccen = 5.439 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 19.575 kN.m
My = max(My,My_MinEccen) + MuaddY = 24.982 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 1.27/221.55 = 6 mm
Actual eccenY = My / P = 2.54/221.55 = 11 mm
eccenXMin = (L/500) + (D/30) = (7275/500) + (450/30) = 30 mm
eccenXMin = max(30,20)= 30 mm
eccenYMin = (L/500) + (B/30) = (7275/500) + (300/30) = 25 mm
eccenYMin = max(25,20) = 30 mm
eccenX = max(Actual eccenX,eccenXMin) = max(6,30) = 30 mm
eccenY = max(Actual eccenY,eccenYMin) = max(11,25) = 25 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(30 mm) > 0.05 x 300(15 mm)
and eccenY(25 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00250 345.09 8.92 336.17 76.04 179 13.61
2 226 0.00055 110.28 4.24 106.04 23.99 90 2.15
3 226 -0.00140 -279.23 0.00 -279.23 -63.16 0 0.00
4 226 -0.00334 -356.91 0.00 -356.91 -80.73 -90 7.23
5 226 -0.00529 -360.90 0.00 -360.90 -81.63 -179 14.61
Total-125.50 37.60
xux = 161 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 161/450 = 0.129
Puc = 0.129 x 20.00 x 300 x 450
= 347.41 kN
Pux1 = Puc + Pus(Total)
= 347.41 + (-125.50)
= 221.91 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 161/450
= 0.149
Muc = 347.41 x (0.5 x 450 - 0.149 x 450) = 54.92 kN-m
Mux1 = Muc + Mus(Total)
= 54.92 + (37.60)
= 92.52 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00152 296.58 8.41 288.17 162.96 104 16.95
2 565 -0.00742 -360.90 0.00 -360.90 -204.08 -104 21.22
Total-41.13 38.17
xuy = 81 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 81/300 = 0.098
Puc = 0.098 x 20.00 x 450 x 300
= 263.88 kN
Puy1 = Puc + Pus(Total)
= 263.88 + (-41.13)
= 222.76 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 81/300
= 0.113
Muc = 263.88 x (0.5 x 300 - 0.113 x 300) = 30.64 kN-m
Muy1 = Muc + Mus(Total)
= 30.64 + (38.17)
= 68.81 kN-m
Pu/Puz = 0.142
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.142, an = 1.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((19.58/92.52)1.000) + ((24.98/68.81)1.000)
= 0.575 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG72 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG73 :

General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 7275 mm
From analysis results,loads on column
Axial load,(P) = 197.03 kN
MomentX,(Mx) = 2.40 kN-m
MomentY,(My) = 10.70 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C10.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 7275 x 1.000 ) / 450 = 16.17 > 12.00,
column is slender in this direction.
Puz = (0.45 x fck x Ag) + (0.75 x fy - 0.45 x fck) x Ast
= 0.45 x 20.00 x 135000.01 + (0.75 x 415.00 - 0.45 x 20.00 ) x 1130.97 = 1556836.792 N
k1 = 0.206511
k2 = 0.080400 MPa
Pub = (k1 x fck x Ag) + (k2 x Ast x 100)
= (0.206511 x 20.00 x 135000.01) + (0.08 x 1130.97 x 100) = 566673.035 N
k = (Puz - Pu) / (Puz - Pub) <= 1
= (1556836.792 - 197032.639) / (1556836.792 - 566673.035) = 1.373312
k = 1.373312 > 1 hence k = 1
MuaddX = Pu x (Depth/2000) x (Lo/Depth) x (Lo/Depth) x k
= 197032.64 x (450.00/2000) x (7275.00/450.00) x (7275.00/450.00) x 1.000000
= 11586750.984 N.mm = 11.587 kN.m
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 7275 x 1.000 ) / 300 = 24.25 > 12.00,
column is slender in this direction.
Puz = (0.45 x fck x Ag) + (0.75 x fy - 0.45 x fck) x Ast
= 0.45 x 20.00 x 135000.01 + (0.75 x 415.00 - 0.45 x 20.00 ) x 1130.97 = 1556836.792 N
k1 = 0.195200
k2 = 0.041467 MPa
Pub = (k1 x fck x Ag) + (k2 x Ast x 100)
= (0.195200 x 20.00 x 135000.01) + (0.04 x 1130.97 x 100) = 531729.797 N
k = (Puz - Pu) / (Puz - Pub) <= 1
= (1556836.792 - 197032.639) / (1556836.792 - 531729.797) = 1.326500
k = 1.326500 > 1 hence k = 1
MuaddY = Pu x (Width/2000) x (Lo/Width) x (Lo/Width) x k
= 197032.64 x (300.00/2000) x (7275.00/300.00) x (7275.00/300.00) x 1.000000
= 17380125.046 N.mm = 17.380 kN.m
Mx_MinEccen = 5.822 kN.m
My_MinEccen = 4.837 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 17.409 kN.m
My = max(My,My_MinEccen) + MuaddY = 28.078 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 2.40/197.03 = 12 mm
Actual eccenY = My / P = 10.70/197.03 = 54 mm
eccenXMin = (L/500) + (D/30) = (7275/500) + (450/30) = 30 mm
eccenXMin = max(30,20)= 30 mm
eccenYMin = (L/500) + (B/30) = (7275/500) + (300/30) = 25 mm
eccenYMin = max(25,20) = 30 mm
eccenX = max(Actual eccenX,eccenXMin) = max(12,30) = 30 mm
eccenY = max(Actual eccenY,eccenYMin) = max(54,25) = 54 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(30 mm) > 0.05 x 300(15 mm)
and eccenY(54 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00246 344.06 8.92 335.14 75.81 179 13.57
2 226 0.00043 86.83 3.45 83.38 18.86 90 1.69
3 226 -0.00159 -303.00 0.00 -303.00 -68.54 0 0.00
4 226 -0.00362 -359.29 0.00 -359.29 -81.27 -90 7.27
5 226 -0.00564 -360.90 0.00 -360.90 -81.63 -179 14.61
Total-136.77 37.14
xux = 155 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 155/450 = 0.124
Puc = 0.124 x 20.00 x 300 x 450
= 334.13 kN
Pux1 = Puc + Pus(Total)
= 334.13 + (-136.77)
= 197.35 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 155/450
= 0.143
Muc = 334.13 x (0.5 x 450 - 0.143 x 450) = 53.68 kN-m
Mux1 = Muc + Mus(Total)
= 53.68 + (37.14)
= 90.82 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00140 280.50 8.12 272.37 154.02 104 16.02
2 565 -0.00808 -360.90 0.00 -360.90 -204.08 -104 21.22
Total-50.06 37.24
xuy = 77 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 77/300 = 0.092
Puc = 0.092 x 20.00 x 450 x 300
= 248.70 kN
Puy1 = Puc + Pus(Total)
= 248.70 + (-50.06)
= 198.64 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 77/300
= 0.106
Muc = 248.70 x (0.5 x 300 - 0.106 x 300) = 29.36 kN-m
Muy1 = Muc + Mus(Total)
= 29.36 + (37.24)
= 66.61 kN-m
Pu/Puz = 0.127
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.127, an = 1.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((17.41/90.82)1.000) + ((28.08/66.61)1.000)
= 0.613 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG73 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG74 :

General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 7275 mm
From analysis results,loads on column
Axial load,(P) = 196.69 kN
MomentX,(Mx) = 2.41 kN-m
MomentY,(My) = 9.84 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C20.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 7275 x 1.000 ) / 450 = 16.17 > 12.00,
column is slender in this direction.
Puz = (0.45 x fck x Ag) + (0.75 x fy - 0.45 x fck) x Ast
= 0.45 x 20.00 x 135000.01 + (0.75 x 415.00 - 0.45 x 20.00 ) x 1130.97 = 1556836.792 N
k1 = 0.206511
k2 = 0.080400 MPa
Pub = (k1 x fck x Ag) + (k2 x Ast x 100)
= (0.206511 x 20.00 x 135000.01) + (0.08 x 1130.97 x 100) = 566673.035 N
k = (Puz - Pu) / (Puz - Pub) <= 1
= (1556836.792 - 196687.210) / (1556836.792 - 566673.035) = 1.373661
k = 1.373661 > 1 hence k = 1
MuaddX = Pu x (Depth/2000) x (Lo/Depth) x (Lo/Depth) x k
= 196687.21 x (450.00/2000) x (7275.00/450.00) x (7275.00/450.00) x 1.000000
= 11566437.721 N.mm = 11.566 kN.m
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 7275 x 1.000 ) / 300 = 24.25 > 12.00,
column is slender in this direction.
Puz = (0.45 x fck x Ag) + (0.75 x fy - 0.45 x fck) x Ast
= 0.45 x 20.00 x 135000.01 + (0.75 x 415.00 - 0.45 x 20.00 ) x 1130.97 = 1556836.792 N
k1 = 0.195200
k2 = 0.041467 MPa
Pub = (k1 x fck x Ag) + (k2 x Ast x 100)
= (0.195200 x 20.00 x 135000.01) + (0.04 x 1130.97 x 100) = 531729.797 N
k = (Puz - Pu) / (Puz - Pub) <= 1
= (1556836.792 - 196687.210) / (1556836.792 - 531729.797) = 1.326837
k = 1.326837 > 1 hence k = 1
MuaddY = Pu x (Width/2000) x (Lo/Width) x (Lo/Width) x k
= 196687.21 x (300.00/2000) x (7275.00/300.00) x (7275.00/300.00) x 1.000000
= 17349655.151 N.mm = 17.350 kN.m
Mx_MinEccen = 5.812 kN.m
My_MinEccen = 4.829 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 17.379 kN.m
My = max(My,My_MinEccen) + MuaddY = 27.192 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 2.41/196.69 = 12 mm
Actual eccenY = My / P = 9.84/196.69 = 50 mm
eccenXMin = (L/500) + (D/30) = (7275/500) + (450/30) = 30 mm
eccenXMin = max(30,20)= 30 mm
eccenYMin = (L/500) + (B/30) = (7275/500) + (300/30) = 25 mm
eccenYMin = max(25,20) = 30 mm
eccenX = max(Actual eccenX,eccenXMin) = max(12,30) = 30 mm
eccenY = max(Actual eccenY,eccenYMin) = max(50,25) = 50 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(30 mm) > 0.05 x 300(15 mm)
and eccenY(50 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00246 344.06 8.92 335.14 75.81 179 13.57
2 226 0.00043 86.83 3.45 83.38 18.86 90 1.69
3 226 -0.00159 -303.00 0.00 -303.00 -68.54 0 0.00
4 226 -0.00362 -359.29 0.00 -359.29 -81.27 -90 7.27
5 226 -0.00564 -360.90 0.00 -360.90 -81.63 -179 14.61
Total-136.77 37.14
xux = 155 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 155/450 = 0.124
Puc = 0.124 x 20.00 x 300 x 450
= 334.13 kN
Pux1 = Puc + Pus(Total)
= 334.13 + (-136.77)
= 197.35 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 155/450
= 0.143
Muc = 334.13 x (0.5 x 450 - 0.143 x 450) = 53.68 kN-m
Mux1 = Muc + Mus(Total)
= 53.68 + (37.14)
= 90.82 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00140 280.50 8.12 272.37 154.02 104 16.02
2 565 -0.00808 -360.90 0.00 -360.90 -204.08 -104 21.22
Total-50.06 37.24
xuy = 77 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 77/300 = 0.092
Puc = 0.092 x 20.00 x 450 x 300
= 248.70 kN
Puy1 = Puc + Pus(Total)
= 248.70 + (-50.06)
= 198.64 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 77/300
= 0.106
Muc = 248.70 x (0.5 x 300 - 0.106 x 300) = 29.36 kN-m
Muy1 = Muc + Mus(Total)
= 29.36 + (37.24)
= 66.61 kN-m
Pu/Puz = 0.126
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.126, an = 1.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((17.38/90.82)1.000) + ((27.19/66.61)1.000)
= 0.600 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG74 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.




Design of CG75 :

General Design Parameters :
Below Floor2 - Default Level at 7.800 m
Column size, (B x D)=300 x 450 mm                Column height,(L)= 7275 mm
From analysis results,loads on column
Axial load,(P) = 212.08 kN
MomentX,(Mx) = 1.26 kN-m
MomentY,(My) = 1.83 kN-m
fck=20.00 N/mm˛
fy= 415.00 N/mm˛
Load combination = 1.50 DL + 1.50 LL
Column NameOrientation Angle
C70.00

Check For Slenderness :
Slenderness RatioX = ( Lo x Effective Length FactorX ) / Depth
= ( 7275 x 1.000 ) / 450 = 16.17 > 12.00,
column is slender in this direction.
Puz = (0.45 x fck x Ag) + (0.75 x fy - 0.45 x fck) x Ast
= 0.45 x 20.00 x 135000.01 + (0.75 x 415.00 - 0.45 x 20.00 ) x 1130.97 = 1556836.792 N
k1 = 0.206511
k2 = 0.080400 MPa
Pub = (k1 x fck x Ag) + (k2 x Ast x 100)
= (0.206511 x 20.00 x 135000.01) + (0.08 x 1130.97 x 100) = 566673.035 N
k = (Puz - Pu) / (Puz - Pub) <= 1
= (1556836.792 - 212080.353) / (1556836.792 - 566673.035) = 1.358115
k = 1.358115 > 1 hence k = 1
MuaddX = Pu x (Depth/2000) x (Lo/Depth) x (Lo/Depth) x k
= 212080.35 x (450.00/2000) x (7275.00/450.00) x (7275.00/450.00) x 1.000000
= 12471651.077 N.mm = 12.472 kN.m
Slenderness RatioY = ( Lo x Effective Length FactorY ) / Width
= ( 7275 x 1.000 ) / 300 = 24.25 > 12.00,
column is slender in this direction.
Puz = (0.45 x fck x Ag) + (0.75 x fy - 0.45 x fck) x Ast
= 0.45 x 20.00 x 135000.01 + (0.75 x 415.00 - 0.45 x 20.00 ) x 1130.97 = 1556836.792 N
k1 = 0.195200
k2 = 0.041467 MPa
Pub = (k1 x fck x Ag) + (k2 x Ast x 100)
= (0.195200 x 20.00 x 135000.01) + (0.04 x 1130.97 x 100) = 531729.797 N
k = (Puz - Pu) / (Puz - Pub) <= 1
= (1556836.792 - 212080.353) / (1556836.792 - 531729.797) = 1.311821
k = 1.311821 > 1 hence k = 1
MuaddY = Pu x (Width/2000) x (Lo/Width) x (Lo/Width) x k
= 212080.35 x (300.00/2000) x (7275.00/300.00) x (7275.00/300.00) x 1.000000
= 18707475.662 N.mm = 18.707 kN.m
Mx_MinEccen = 6.267 kN.m
My_MinEccen = 5.207 kN.m
Mx = max(Mx,Mx_MinEccen) + MuaddX = 18.739 kN.m
My = max(My,My_MinEccen) + MuaddY = 23.914 kN.m

Calculation Of Eccentricities :
              As per IS 456: 2000 Clause 25.4,all columns shall be designed for minimum eccentricity
equal to the unsupported length of the column/500 plus lateral dimension/30, subject to a minimum of 20 mm.
Actual eccenX = Mx / P = 1.26/212.08 = 6 mm
Actual eccenY = My / P = 1.83/212.08 = 9 mm
eccenXMin = (L/500) + (D/30) = (7275/500) + (450/30) = 30 mm
eccenXMin = max(30,20)= 30 mm
eccenYMin = (L/500) + (B/30) = (7275/500) + (300/30) = 25 mm
eccenYMin = max(25,20) = 30 mm
eccenX = max(Actual eccenX,eccenXMin) = max(6,30) = 30 mm
eccenY = max(Actual eccenY,eccenYMin) = max(9,25) = 25 mm
As per IS 456: 2000 Clause 39.3,
when the minimum eccentricity does not exceed 0.05 times the lateral dimension,
the column will be designed as an axially loaded column.
eccenX(30 mm) > 0.05 x 300(15 mm)
and eccenY(25 mm) > 0.05 x 450(23 mm)
               Hence the column will be designed as a column with biaxial moments.
Provided steel :
Provide #12 - 10 nos. (1131 mm˛)
Biaxial Check Calculations :

For X axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 226 0.00249 344.80 8.92 335.88 75.98 179 13.60
2 226 0.00052 103.77 4.03 99.74 22.56 90 2.02
3 226 -0.00145 -289.67 0.00 -289.67 -65.52 0 0.00
4 226 -0.00342 -357.57 0.00 -357.57 -80.88 -90 7.24
5 226 -0.00539 -360.90 0.00 -360.90 -81.63 -179 14.61
Total-129.50 37.47
xux = 159 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 159/450 = 0.127
Puc = 0.127 x 20.00 x 300 x 450
= 343.62 kN
Pux1 = Puc + Pus(Total)
= 343.62 + (-129.50)
= 214.12 kN > P ,hence O.K.
Muc = Puc.(0.5.D - C2.D)
C2 = 0.416 ku = 0.416 x 159/450
= 0.147
Muc = 343.62 x (0.5 x 450 - 0.147 x 450) = 54.57 kN-m
Mux1 = Muc + Mus(Total)
= 54.57 + (37.47)
= 92.04 kN-m

For Y axis :
Row No.
(i)
Asi
(mm˛)
ei
fsi
(N/mm˛)
fci
(N/mm˛)
(fsi - fci)
(N/mm˛)
Pusi
(kN)
xi
(mm)
Musi
(kN-m)
1 565 0.00146 291.03 8.28 282.75 159.89 104 16.63
2 565 -0.00774 -360.90 0.00 -360.90 -204.08 -104 21.22
Total-44.19 37.85
xuy = 79 mm               Puc = C1.fck.B.D
C1 = 0.36 ku = 0.36 x 79/300 = 0.095
Puc = 0.095 x 20.00 x 450 x 300
= 256.29 kN
Puy1 = Puc + Pus(Total)
= 256.29 + (-44.19)
= 212.10 kN > P ,hence O.K.
Muc = Puc.(0.5.B - C2.B)
C2 = 0.416 ku = 0.416 x 79/300
= 0.110
Muc = 256.29 x (0.5 x 300 - 0.110 x 300) = 30.01 kN-m
Muy1 = Muc + Mus(Total)
= 30.01 + (37.85)
= 67.86 kN-m
Pu/Puz = 0.136
For Pu/Puz <= 0.2, an = 1.0
For Pu/Puz >= 0.8, an = 2.0
For Pu/Puz = 0.136, an = 1.000
((Mux / Mux1)an) + ((Muy/Muy1)an)
= ((18.74/92.04)1.000) + ((23.91/67.86)1.000)
= 0.556 < 1.0,hence O.K.

Ties Details :
Load combination for ties design = 1.50 DL + 1.50 LL
Calculation of diameter of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the diameter of the polygonal links or
lateral ties shall be not less than the following
               i) one-fourth the diameter of the largest longitudinal bar = 12/4 = 3 mm
               ii) in no case less than = 6 mm
Required diameter = maximum of (3, 6) = 6 mm
               Provide 8 mm diameter for ties.

Calculation of spacing of ties :
As per IS 456: 2000 Clause 26.5.3.2 (c), the pitch of transverse reinforcement shall be
               not more than the least of the following distances:
               i) the least lateral dimension of the compression member = 300 mm
               ii) sixteen times the smallest diameter of the longitudinal reinforcement bar to be tied
               = 16 x 12 = 192 mm
               iii) 300 mm
               Required spacing = minimum of (300, 192, 300) = 192 mm
# Provide ties               8 @ 190 mm c/c

SUMMARY :

CG75 : Floor2 - Default Level at 7.800 m
Provide rectangular section : 300 x 450 mm
Provide #12 - 10 nos. (1131 mm˛)
Provide ties #8 @ 190 mm c/c
Provide 4 legged ties along width and
2 legged ties along depth.